339. Derivatives. Rolle's function.

by Virgil Nicula, Apr 5, 2012, 5:44 PM

I. Derivatives.

$3.1.\blacktriangleright$ Prove that $f(x)=x\sqrt{x+1}\implies$ $f^{(n)} (x)=\frac{(-1)^n(2n-5)!!}{2^n}\cdot \frac{(3x+2n)\sqrt {x+1}}{(x+1)^{n-1}}$ .

$3.2.\blacktriangleright$ $(\arctan)^{(n)}=(-1)^{n-1}(n-1)!\cdot \frac {\sin\left(n\cdot\arctan\frac 1x\right)}{\sqrt {\left(x^2+1\right)^n}}$ .

$3.3.\blacktriangleright$ $\left[\ln \left(x^2+1\right)\right]^{(n)}=2(-1)^n(n-1)!\cdot\frac {\cos\left(n\cdot \arctan\frac 1x\right)}{\sqrt {\left(x^2+1\right)^n}}$ .

$3.4.\blacktriangleright$ For $\Delta\equiv a^2-4b<0$ prove that $\left(\frac {1}{x^2+ax+b}\right)^{(n)}=\frac {2(-1)^nn!}{\sqrt {4b-a^2}}\cdot\frac {\sin\left[(n+1)\arctan\frac {\sqrt {4b-a^2}}{2x+a}\right]}{\sqrt {\left(x^2+ax+b\right)^{n+1}}}$ .

$3.5.\blacktriangleright$ $\left\{\begin{array}{c}n\in \mathrm N^{*}\ ;\ x_{k}\in\mathrm R\ ,\ k\in \overline{1,n}\\\ na=x_{1}+x_{2}+\ldots+x_{n}\\\ nb=x^{2}_{1}+x^{2}_{2}+\ldots+x^{2}_{n}\end{array}\right\|\Longrightarrow\left(\forall\right) k\in\overline{1,n}\ ,\ \left|x_{k}-a\right|\le\sqrt{(n-1)\left(b-a^{2}\right)}$ .

For example, $\left\{\begin{array}{c}\{a,b,c\}\subset \mathrm R\\\\ a+b+c=2\\\\ ab+bc+ca=1\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}
\{a,b,c\}\subset \left[0,\frac 43\right]\\\\ 
abc\le \frac{4}{27}\end{array}\right\|$.

$3.6.\blacktriangleright$ Let $x_{i}$ be real numbers, with $i \in \overline{1,n}$. Prove that $\frac{\sum{\left( x_{i}-\frac{\sum{x_{i}}}{n}\right)^{2}}}{n}\leq  \left(\frac{\max{x_{i}}-\min{x_{i}}}{2}\right)^{2}$ (see here).

$3.7.\blacktriangleright$ Let $f$ be a function $f(x)=a_0x^3+a_1x^2+a_2x+a_3,\ a_0\ne 0,\ x\in \mathbb R$ and a line $d$ for which $ d\cap {G_f} =\left\{P_k |k\in \overline {1,3}\right\}$ .


For any $k\in\overline{1,3}$ the tangent at $G_f$ in the point $P_k\in {G_f}$ meets again the graph $G_f$ in the point $Q_k$ . Prove that the point $Q_k,\ k\in \overline{1,3}$


are collinear. In particular case, the extremum points (if exist) and the inflexion point of $G_f$ are colinearly.


Proof. Prove easily that the points $X_k(x_k,f(x_k)),\ k\in \overline{1,3}$ are collinear$\Longleftrightarrow$ $x_1+x_2+x_3=-\frac{a_1}{a_0}$ . The points

$P_k,\ k\in \overline {1,3}$ are collinear $\Longleftrightarrow p_1+p_2+p_3=-\frac{a_1}{a_0}. $ But the points $P_k,P_k,Q_k $ are collinear, i.e. $2p_k+q_k=-\frac{a_1}{a_0},\ k\in \overline {1,3}$ .

Thus, $2(p_1+p_2+p_3)+(q_1+q_2+q_3)=-\frac{3a_1}{a_0}$ $\iff$ $- \frac{2a_1}{a_0}+(q_1+q_2+q_3)=-\frac{3a_1}{a_0}$ $\iff$ $q_1+q_2+q_3=-\frac{a_1}{a_0}$ $\iff$ the points $Q_1,Q_2,Q_3$ are collinear.


$3.8.\blacktriangleright$ Ascertain the range of the function $f(x,y)\equiv\sin x+\sin y-\sin x\cdot \sin y$ , where $0<x,y\le \frac{\pi}{2}\le x+y<\pi $ .

$3.9.\blacktriangleright$ Prove that for any $\{x,y,z\}\subset (0,\infty )$ exists the inequality $\left[x^2(y+z)\right]^{y-x}\cdot\left[y^2(z+x)\right]^{z-y}\cdot\left[z^2(x+y)\right]^{x-z}\le 1$ .



Proof. Consider the function $f:(0,\infty )\rightarrow \mathbb R$ , where $f(x)=\frac {x^y}{(x+y)^{y+z}}$ . Observe that $f'(x)\ .s.s.\ y^2-xz$ . Therefore, $\frac {x^y}{(x+y)^{y+z}}=$ $f(x)\le f\left(\frac {y^2}{z}\right)=$ $\frac {z^zy^y}{y^z(y+z)^{y+z}}$ , i.e.

$\left(\frac {y+z}{x+y}\right)^{y+z}\le \left(\frac yx\right)^y\cdot\left(\frac zy\right)^z$ . From circular permutations obtain still two relations. From the product of these three inequalities results the proposed inequality.



II. Exercises with Rolle's functions.

$4.1.\blacktriangleright$ Consider a continue function $f : [a,b]\rightarrow (-1,1)$ so that $f(a)f(b)<0$ . Prove that exists $c\in (a,b)$ such that $f(c)=\frac {f(a)+f(b)}{1+f(a)f(b)}$ .

Proof. Let $h(x)=f(x)[1+f(a)f(b)]-[f(a)+f(b)]\ ,\ a\le x\le b$ . Observe that $h$ is continue and $h(a)h(b)=f(a)f(b)\left[f^2(a)-1\right]\left[f^2(b)-1\right]<0$

because $f(a)f(b)<0$ and $\left\{f(a),f(b)\right\}\subset (-1,1)$ . Therefore, there is $c\in (a,b)$ so that $h(c)=0$ , i.e. $f(c)=\frac{f(a)+f(b)}{1+f(a)f(b)}$ .


$4.2.\blacktriangleright$ Consider a Rolle's function $f : [0,1]\rightarrow \mathbb R$ for which $f(1)=0$ . Prove that $\forall m\ne 0,\ \exists\ c_m\in (0,1)$ such that $c^2_m\cdot f'(c_m)+|m|\cdot f(c_m)=0$ .[/quote]

Proof. Construct the function $G:[0,1]\rightarrow \mathbb R$ , $G(0)=0$ and $G(x)=f(x)\cdot e^{-\frac{|m|}{x}},\ x\in (0,1]$ . Observe that $G(0+0)=\lim_{x\searrow 0} f(x)\cdot e^{-\frac{|m|}{x}}=0$ . Thus

the function $G$ is a Rolle's function and $G(0)=G(1)=0$ . Results that there is $c\in (0,1)$ such that $G'(c)=0$ , i.e. $c^2\cdot f'(c)+|m|\cdot f(c)=0$ .


$4.3.\blacktriangleright$ Let $f:[a,b]\rightarrow \mathbb R$ be a Rolle's function , where $0<a<b$ .
Prove that there are $c<d,$ , $\{c,d\}\subset (a,b)$ such that $c(c-a)f'(c)=a(b-c)f'(d)$ .


Proof. Construct the function $F:[a,b]\rightarrow\mathbb R$ , where $F(x)=\frac {(x-a)[f(a)-f(x)]}{x}$ . Observe that $F$ is Rolle's function and $F(a)=F(b)=0$ . From Rolle's theorem obtain that

exists $c\in (a,b)$ so that $F'(c)=0$ , i.e. $c(a-c)f'(c)=a[f(c)-f(b)]$ . Also exists $d\in (c,b)$ so that $f(c)-f(b)=f'(d)(c-b)$ a.s.o.


$4.4.\blacktriangleright$ Prove that for any $n\in \mathbb N$ , $n\ge 2$ the equation $\sqrt [n]{x+1}-\sqrt [n] x=\frac{1}{n}$ has only one zero $x_{n}\in (0,1)$ . Prove that $\lim_{n\to\infty}x_{n}=\frac {1}{e-1}$ .

Proof. Denote $f_n:[0,1]\rightarrow\mathbb R$ , where $f(x)=\sqrt [n]{x+1}-\sqrt [n] x-\frac{1}{n}$ . Prove easily that $f'(x)\ .a.s.\ \sqrt [n]{x^{n-1}}-\sqrt [n]{(x+1)^{n-1}}\ .a.s.\ x-(x+1)<0$ . Thus

the function $f$ is decreasing on $(0,1]$ and $f_n(0)=1-\frac 1n>0>\sqrt [n]2-\left(1+\frac 1n\right)=f_n(1)$ . Therefore exists and is unique $x_n\in (0,1)$ so that $f_n\left(x_n\right)=0$ .

Denote $g_n(x)=n\cdot\left(\sqrt [n]{x+1}-\sqrt [n] x\right)-1$ and $h(x)=\ln\frac {x+1}{x}-1$ , where $x\in (0,1)$ . Observe that $f(x)\ .a.s.\ g(x)$ , $\lim_{n\to\infty}g_n(x)=h(x)$ , $h(x)=0\iff x=l$ ,

where $l=\frac {1}{e-1}$ and $h(x)\ .s.s.\ (l-x)$ for any $x\in (0,1)$ . Let $U=(\alpha , \beta )$ be a neighbourhood of $l$ . Observe that

$:\ \left\{\begin{array}{ccc}
\lim_{n\to\infty}g_n(\alpha )=\ln\frac {\alpha +1}{\alpha }-1>0 & \implies & (\exists)\ n_1\in\mathbb N^*\ (\forall )n\ge n_1\ ,\ g_n(\alpha )>0\\\\
\lim_{n\to\infty}g_n(\beta )=\ln\frac {\beta +1}{\beta }-1>0 & \implies & (\exists)\ n_2\in\mathbb N^*\ (\forall )n\ge n_2\ ,\ g_n(\beta )<0\end{array}\right\|$ . Thus exists $p\in\mathbb N^*$ , $p=\max\left\{n_1,n_2\right\}$ so that for any $n>p$ we have

$g_n(\beta)<0=g_n(x_n)<g_n(\alpha )$ $\implies$ $x_n\in\left(\alpha , \beta \right)$ . For any neighbourhood $U=(\alpha , \beta )$ of $l$ exists $p\in \mathbb N^*$ so that for $n\ge p$ we have $x_n\in U$ , i.e. $\lim_{n\to\infty}x_n=l=\frac {1}{e-1}$ .


$4.5.\blacktriangleright$ Consider a Rolle's function $f : [a,b]\rightarrow \mathbb R$ for which $f(a)=f(b)$ . Prove that for any $n\in\mathbb N^*$ exists $x_n\in [a,b]$ so that $f(x_n)=f\left(x_n+\frac {b-a}{n}\right)$ .

$4.6.\blacktriangleright$ Given is the function $f: [a,b]\rightarrow \mathbb R$ so that $f\in C^1$ , $f'(a)=f'(b)$.
Prove that exists $c\in (a,b)$ so that $(b-c)\left[f'(c)-\frac{f(c)-f(a)}{c-a}\right]=(c-a)\left[f'(c)-\frac{f(b)-f(c)}{b-c}\right].$

$4.7.\blacktriangleright$ Given is the function $f: [a,b]\rightarrow \mathbb R$ so that $f\in C^1$ for which exists $m\ne n$ so that $\frac{mf'(a)-nf'(b)}{m-n}=\frac{f(a)-f(b)}{a-b}$.
Prove that exists $c\in (a,b)$ such that $m(b-c)\left[f'(c)-\frac{f(c)-f(a)}{c-a}\right]=n(c-a)\left[f'(c)-\frac{f(b)-f(c)}{b-c}\right]$ .

$4.8.\blacktriangleright$ Given are $a$ , $b$ so that $a>1$ , $b>1$ and $c=\sqrt {\log_ab}$ . Prove that the function $f:[c,\infty )\rightarrow \mathbb R$ ,
where $f(x)=a^x+b^{\frac 1x}$ is strict increasing and solve the equation $x+16^{\log_x2}=18$ .

$4.9.\blacktriangleright$ Solve in $\mathbb R$ the system of equations $\left\{\begin{array}{c}
e^x+e^y=2\\\
x+y=xy\end{array}\right\|$ .

$4.10.\blacktriangleright$ Consider the function $f(x)=\sqrt [3]{x^3+ax^2+1}+\sqrt [3]{x^3-ax^2+1}-2x$ , where $x\in \mathbb R$ and $a>0$ .

Prove that exists $b>0$ so that $f(x)<0$ for any $x>b$ and the equation $f(x)=0$ has at least one zero for any $a\in\left(0,\frac {10}{3\sqrt 3}\right)$ .

$4.11.\blacktriangleright$ Consider the function $f:[a,b]\rightarrow\mathbb R$ , where $0<a<b$ and $f(x)=\ln x$ . Prove that the point $c\in (a,b)$ which given by the

Lagrange's theorem is uniquely and $c\in\left(\sqrt {ab},\frac {a+b}{2}\right)\subset (a,b)$ . Deduct the inequalities $\left\{\begin{array}{cc}
(1) & \frac {2}{2t+1}<\ln \left(1+\frac 1t\right)<\frac {1}{\sqrt{t(t+1)}}\ (\forall )\ t>0\ .\\\\
(2) & 2\left(e^{\pi}-e^e\right)<(\pi -e)\left(e^{\pi}+e^e\right)\ .\end{array}\right\| . $


Proof. The point $c\in (a,b)$ is given by relation $\frac {\ln b-\ln a}{b-a}=\frac 1c$ .Must prove $\sqrt {ab}<\frac {b-a}{\ln\frac ba}<\frac {a+b}{2}$ , i.e. $\boxed{2\cdot\frac {b-a}{b+a}<\ln \frac ba<\frac {b-a}{\sqrt {ab}}}\ (*)$ .

Denote $\frac ba=t>1$ . Thus the relation $(*)$ becomes $2\cdot\frac {t-1}{t+1}<\ln t<\frac {t-1}{\sqrt t}$ , where $t>1$ . Therefore,

$\left\|\begin{array}{ccccccc}
h(t)\equiv\ln t-2\cdot\frac {t-1}{t+1} & \implies & h'(t)=\frac {(t-1)^2}{t(t+1)^2}> 0 & \implies & h(t)>h(1) & \implies & h(t)>0\ .\\\\
g(t)\equiv\frac {t-1}{\sqrt t}-\ln t & \implies & h'(t)=\frac {\left(\sqrt t-1\right)^2}{2t\sqrt t}> 0 & \implies & g(t)>g(1) & \implies & g(t)>0\ .\end{array}\right\|$

Can obtain the inequalities $(1)$ , $(2)$ from $(*)$ thus : for the first put $b:=t+1$ , $a:=t$ , where $t>0$ ; for the second put $a:=e^e$ , $b:=e^{\pi}$ .



$4.12.\blacktriangleright$ Prove that $\left(\frac {x}{\sin x}\right)^{\tan x-x}\ >\ \left(\frac {\tan x}{x}\right)^{x-\sin x}$ for any $x\in\left(0,\frac {\pi}{2}\right)$ .

Proof. Since $0<\sin x<x<\tan x$ obtain $\left(\frac {x}{\sin x}\right)^{\tan x-x}\ >\ \left(\frac {\tan x}{x}\right)^{x-\sin x}$ for any $x\in\left(0,\frac {\pi}{2}\right)$ $\iff$ $\frac {\ln x-\ln\sin x}{x-\sin x}>\frac {\ln\tan x-\ln x}{\tan x-x}$

for any $x\in\left(0,\frac {\pi}{2}\right)$ . Apply the Lagrange's theorem to the function $\ln\ :\ (0,\infty )\rightarrow\mathbb R$ on the segments $[\sin x,x]$ and $[x,\tan x]$ .

Therefore, exist $0<\sin x<c_x<x<d_x<\tan x$ so that $\frac {\ln x-\ln\sin x}{x-\sin x}=\frac {1}{c_x}$ and $\frac {\ln\tan x-\ln x}{\tan x-x}=\frac {1}{d_x}$ . In conclusion,

$0<c_x<d_x$ $\iff$ $\frac {1}{c_x}>\frac {1}{d_x}$ $\iff$ $\frac {\ln x-\ln\sin x}{x-\sin x}>\frac {\ln\tan x-\ln x}{\tan x-x}$ $\iff$ $\left(\frac {x}{\sin x}\right)^{\tan x-x}\ >\ \left(\frac {\tan x}{x}\right)^{x-\sin x}$ for any $x\in\left(0,\frac {\pi}{2}\right)$ .



$4.13.\blacktriangleright$ Let $P\in \mathbb R[X]$ be a polynomial with at least two roots $a<b$ . Prove that for any real number $k$ there is $c\in (a,b)$ such that $P(c)+kP'(c)=0$ .

Proof. If $k=0$ , then can choose $c\in\{a,b\}$ . Suppose $k\ne 0$ . In this case consider $f:[a,b]\rightarrow \mathbb R$ , where $f(x)=e^{\frac xk}\cdot P(x)$ .

Observe that $f$ is a Rolle function , i.e. a continue function in $a$ , $b$ and a differentiable function on $(a,b)$ . Since $f(a)=f(b)=0$ and

$f'(x)=e^{\frac xk}\cdot\left[P'(x)+\frac 1k\cdot P(x)\right]$ , from the Rolle's theorem obtain that there is $c\in (a,b)$ so that $f'(c)=0$ , i.e. $P(c)+kP'(c)=0$ .


$4.14\blacktriangleright$ Prove that the sum $s_n=\sum_{k=1}^n\frac {1}{k\sqrt k}$ is convergent (without inegral).

Proof. Consider $f:[k,k+1]\rightarrow\mathbb R$ , where $f(x)=\frac {1}{\sqrt x}$ and $k\in\overline {1,n-1}$ . Observe that $f'(x)=-\frac {1}{2x\sqrt x}$ . From the

Lagrange's theorem results that exists $c_k\in [k,k+1]$ so that $f(k+1)-f(k)=f'(c_k)$ . Since the function $f'$ is increasing obtain

$f'(k)<f(k+1)-f(k)<f'(k+1)$ , i.e. $\frac 1{2k\sqrt k}\ge\frac 1{\sqrt k}-\frac 1{\sqrt {k+1}}\ge \frac {1}{2(k+1)\sqrt {k+1}}\ ,\ \sum_{k=1}^n\ \implies$

$s_n>2\cdot\left(1-\frac {1}{\sqrt {n+1}}\right)>s_n-1+\frac {1}{\sqrt{n+1}}\iff$ $2\cdot\left(1-\frac {1}{\sqrt{n+1}}\right)\ <\ s_n\ <\ 3\cdot\left(1-\frac {1}{\sqrt{n+1}}\right)$ . Our sequence is increasing and bounded $2-\sqrt 2\le 2\cdot\left(1-\frac {1}{\sqrt{n+1}}\right)<s_n<3\cdot\left(1-\frac {1}{\sqrt{n+1}}\right)<3$ obtain $s_n\nearrow L\in [2,3]$ .


PP15. For a given $a\in\left(0,\frac {\pi}{2}\right)$ consider the function $f(x)=\frac{\sin2x}{\cos(x-a)} $ , where $x\in\left(0,\frac{\pi}{2}\right)$ where $a$ . Is it an injective function ? Find $\max_{x\in \left(0,\frac {\pi}{2}\right)}f(x)$ .

Proof. Since $\left\{\begin{array}{c}
f(0)=0\\\\
f(a)=\sin 2a>0\\\\
f\left(\frac {\pi}{2}\right)=0\end{array}\right\|$ and the function $f$ is continue we can apply the Darboux's property: for any $\lambda\in (0,f(a))$ exist $c_1\ne c_2$ for which

$\left\{\begin{array}{ccc}
c_1\in (0,a) & \mathrm{so\ that} & f\left(c_1\right)=\lambda\\\\
c_2\in \left(a,\frac {\pi}{2}\right)  & \mathrm{so\ that} & f\left(c_2\right)=\lambda\end{array}\right\|\implies$ $f\left(c_1\right)=f\left(c_2\right)$ , i.e. our function isn't injectively. Define

$X\ .s.s.\ Y\iff X=Y=0\ \vee\ XY>0$ , i.e. $\mathrm{sgn}(X)=\mathrm{sgn}(Y)$ . So $f'(x)\ .s.s.\ 2\cos 2x\cos (x-a)+$ $\sin (x-a)\sin 2x=$

$\cos 2x\cos (x-a)+\cos (x+a)=$ $\cos 2x(\cos x\cos a+\sin x\sin a)+$ $\cos x\cos a-\sin x\sin a=$ $\cos x\cos a(1+\cos 2x)-\sin x\sin a(1-\cos 2x)=$

$2\cos^3x\cos a-2\sin^3x\sin a\ .s.s.$ $\cot a-\tan^3x\ .s.s.$ $\sqrt [3]{\cot a}-\tan x\ .s.s.$ $\arctan\left(\sqrt [3]{\cot a}\right)-x\implies$ $f'(x)\ .s.s.\ \arctan\left(\sqrt [3]{\cot a}\right)-x$ .

In conclusion, the function $f$ is $\nearrow$ on $\left(0,\arctan\left(\sqrt [3]{\cot a}\right)\right]$ and is $\searrow$ on $\left[\arctan\left(\sqrt [3]{\cot a}\right),\frac {\pi}{2}\right)\implies$ $\boxed{\ \max_{x\in \left(0,\frac {\pi}{2}\right)}f(x)=f\left[\arctan\left(\sqrt [3]{\cot a}\right)\right]\ }$ .



PP16. Find $\min_{a>0}\ I(a)$ , where $I(a)=\int_1^e |\ln ax|\ \mathrm{dx}\ .$

Proof. Let $\int\ln ax\ \mathrm{dx}=g(x)+\mathbb C$ , where $g(x)=x\ln a+x\ln x-x\ ,\ x>0$ . Observe that $\left\{\begin{array}{ccc}
g(1)=\ln a-1\\\\
g(e)=e\ln a\\\\
g\left(\frac 1a\right)=-\frac 1a\end{array}\right\|$ . Define the relation

$X\sim Y\ \iff\ \mathrm{sign}(X)=\mathrm{sign}(Y)$ , i.e. $X=Y=0\ \vee\ XY>0$ .Observe that $\ln ax\sim (ax-1)\sim \left(x-\frac 1a\right)$ . Appear three cases :

$1.\blacktriangleright\ \boxed{\ 0< a< \frac 1e\ }$ , i.e. $e<\frac 1a\implies$ $I(a)=-\int_1^e\ln ax\ \mathrm{dx}=-g(e)+g(1)=$ $-e\ln a+\ln a-1\implies$ $\boxed{\ I(a)= -(e-1)\ln a-1\ }$ .

$2.\blacktriangleright\ \boxed{\ \frac 1e\le a\le 1\ }$ , i.e. $1\le \frac 1a\le e$ and $\ln ax\sim\left(x-\frac 1a\right)$ for any $x\in [1,e]\implies$ $I(a)=-\int_1^{\frac 1a}\ln ax\ \mathrm{dx}+\int_{\frac 1a}^e\ln ax\ \mathrm{dx}=$

$g(1)+g(e)-2\cdot g\left(\frac 1a\right)=$ $(\ln a-1)+e\ln a-2\cdot \left(-\frac 1a\right)\implies$ $\boxed{\ I(a)=\frac 2a+(e+1)\cdot\ln a-1\ }$ . Observe that $(\forall )\ a\in\left(\frac 1e,1\right)$

we have $I'(a)=\left(-\frac {2}{a^2}+\frac {e+1}{a}\right)\sim \left[(e+1)a-2\right]\sim \left(a-\frac {2}{e+1}\right)$ , i.e. $\min_{\frac 1e\le a\le 1}\ I(a)=I\left(\frac {2}{e+1}\right)$ .

$3.\blacktriangleright\ \boxed{\ 1<a\ }$ , i.e. $0<\frac 1a<1<e\implies$ $I(a)=\int_1^e\ln ax\ \mathrm{dx}=$ $g(e)-g(1)=$ $e\ln a-\ln a+1\implies$ $\boxed{\ I(a)=(e-1)\ln a+1\ }$ .

In conclusion, $I(a)=\left\{\begin{array}{cccc}
-(e-1)\ln a-1 & \searrow  & \iff & 0< a< \frac 1e\\\\
\frac 2a+(e+1)\cdot\ln a-1 & \searrow\ \nearrow  & \iff & \frac 1e\le a\le 1\\\\
(e-1)\ln a+1 &  \nearrow & \iff & 1<a\end{array}\right\|$ , where $\left\{\begin{array}{ccc}
I\left(\frac 1e\right) & = & e-2\\\\
I\left(\frac {2}{e+1}\right) & = & (e+1)\cdot\ln\frac {2e}{e+1}-1\\\\
I(1) & = & 1\end{array}\right\|$ .

Therefore, the evolution of the function $I(a)\ ,\ a>0$ is $\begin{array}{cccc}
\searrow & & & \nearrow\\\
& \searrow & \nearrow & \end{array}$ and $\boxed{\ \min_{a>0}\ I(a)=I\left(\frac {2}{e+1}\right)=(e+1)\cdot\ln\frac {2e}{e+1}-1\ }$ .


See here and here
This post has been edited 28 times. Last edited by Virgil Nicula, Nov 26, 2015, 3:45 PM

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For the problem $4.12$ there is a solution that doesn't use Lagrange's theorem :lol: We shall use the following charactherization of convex functions which states that :
$\color{white}{.}$
A function $f$ is convex on the interval $I\subset\mathbb{R}$ if and only if : $\frac {f(a)}{(a-b)(a-c)}+\frac {f(b)}{(b-a)(b-c)}+\frac {f(c)}{(c-a)(c-b)}\ge 0\ ,$ for any distinct $a,b,c\in I$ . If the inequality is strict, then $f$ is

strict convex.
For $a<b<c\subset I$ the above inequality is equivalent to : $\boxed{\ (c-b)f(a)+(a-c)f(b)+(b-a)f(c)\ge 0\ }$ $\iff\ \boxed{\ \frac {f(b)-f(a)}{b-a}\ \le\ \frac {f(c)-f(b)}{c-b}\ }\ \ (\star)$ .

On the interval $\left(\ 0\ ,\ \frac {\pi}2\ \right)$ we need to take $f(x)=-\ln x$ (which is strict convex) and $\left\|\ \begin{array}{ccc}
a & = & \sin x \\  
b & = & x \\ 
c & = & \tan x\ \end{array}\right\|$ in order to obtain the inequality :

$ \frac {\ln x-\ln\sin x}{x-\sin x}\ >\ \frac {\ln\tan x-\ln x}{\tan x-x}$ $\iff\ \boxed{\ \left(\frac x{\sin x}\right)^{\tan x-x}\ >\ \left(\frac {\tan x}x\right)^{x-\sin x}\ }\ \ ,\ \forall\ x\in\left(\ 0\ ,\ \frac {\pi}2\ \right)$ .
This post has been edited 6 times. Last edited by Virgil Nicula, Nov 26, 2015, 3:53 PM

by Mateescu Constantin, Apr 5, 2012, 7:03 PM

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Multumesc.

by Virgil Nicula, Apr 6, 2012, 3:13 PM

Own problems or extensions/generalizations of some problems which was posted here.

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  • We need virgil nicula to return to aops, this blog is top 10 all time.

    by OlympusHero, Sep 14, 2022, 4:44 AM

  • :omighty: blog

    by tigerzhang, Aug 1, 2021, 12:02 AM

  • Amazing blog.

    by OlympusHero, May 13, 2021, 10:23 PM

  • the visits tho

    by GoogleNebula, Apr 14, 2021, 5:25 AM

  • Bro this blog is ripped

    by samrocksnature, Apr 14, 2021, 5:16 AM

  • Holy- Darn this is good. shame it's inactive now

    by the_mathmagician, Jan 17, 2021, 7:43 PM

  • godly blog. opopop

    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404395
  • Total comments: 37
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