181. Two special inequalities from CRUX (own).

by Virgil Nicula, Dec 1, 2010, 10:27 AM

Proposed problem 1. $\{a,b,c\}\subset\mathbb R\ \implies\ \boxed{\ \ a^2\ +\ b^2\ +\ c^2\ =\ 9\ \Longrightarrow\ 3\ \min\ \{\ a\ ,\ b\ ,\ c\ \}\ \le\ 1\ +\ abc\ }$ (CRUX, 3241/4/2007).

Proof. I"ll suppose w.l.o.g. that $a\le b\le c$. Thus, the problem reduces to prove that $3a\le 1+abc$. If $abc=0$ , then $a\le 0$ and $3a< 1+abc$ .

If $a\le b < 0 < c$ , then $abc>0$ and $3a<0<1<1+abc$ , i.e. $3a<1+abc$ . Thus, are remaining only two cases.

$1\blacktriangleright$ If $a\le b\le c<0$ or $a < 0 < b\le c$, then $2bc\le (b^2+c^2)=9-a^2$ and $abc\ge \frac {a(9-a^2)}{2}$ . But $\frac {a(9-a^2)}{2}\ge (3a-1)$ $\iff$ $a^3-3a-2\le 0$ $\iff$

$(a-2)(a+1)^2\le 0$, which is true. Thus, $abc\ge \frac {a(9-a^2)}{2}\ge 3a-1$, i.e. $3a \le 1+abc$. We have equality iff $\boxed{a=-1\ \wedge\ b=c=2}$ .

$2\blacktriangleright$ If $0 < a\le b\le c$, then $9=a^2+b^2+c^2\ge b^2+2a^2$, i.e. $2a^2+b^2-9\le 0$ , $a^2-b^2\le 0$ and $b^2 c^2-a^2\left(9-2a^2\right)=$

$b^2\left(9-a^2-b^2\right)-a^2\left(9-2a^2\right)=$ $\left(a^2-b^2\right)\left(2a^2+b^2-9\right)\ge 0$ , i.e. $b^2 c^2\ge a^2\left(9-2a^2\right)$ . But $a^2\le 3< \frac 92$ $\implies$ $\left\|\begin{array}{c} 9-2a^2> 0\\\\ bc\ge a\sqrt {9-2a^2}\end{array}\right\|$ ,

i.e. $abc\ge a^2\sqrt{9-2a^2}$. Is sufficient to prove $a^2\sqrt{9-2a^2}\ge 3a-1\ (*)$ , i.e. $f(a)=2a^6-9a^4+(3a-1)^2\le 0\ ,\ (\forall )\ a\in \left(0 , \sqrt 3\right]$ .

$2.1\blacktriangleright\ \ :\ \ a\in \left(0, \frac 13 \right]$ $\implies$ $a^2\sqrt{9-2a^2}>0\ge 3a-1$ $\implies$ the relation $(*)$ is truly, i.e. $f(a)\le 0$ .

$2.2\blacktriangleright\ \ :\ \ a\in \left(\frac 13 ,1\right]$ $\implies$ $f(a)=2a^4\left(a^2-1\right)-\left(a^2\sqrt 7 -3a+1\right)\left(a^2\sqrt 7 +3a-1\right)< 0$ .

$2.3\blacktriangleright\ \ :\ \ a\in \left(1, \sqrt{\frac 32}\right]$ $\implies$ $f(a)=\left(a^4+\frac 32\right)\left(2a^2-3\right) - 6a^2\left(a^2-1\right)- 6\left(a-\frac {11}{12}\right)< 0$ .

$2.4\blacktriangleright\ \ :\ \ a\in \left(\sqrt {\frac 32} ,\sqrt 3 \right]$ $\implies$ $f(a)=a^2\left(a^2-3\right)\left(2a^2-3\right)+(1-6a) < 0$ .

Application. Prove that $\boxed{\ \left\{\begin{array}{c} \{a,b,c\}\subset\mathbb R\\\\ a^2 + b^2 + c^2 =9\end{array}\right|\ \implies\ 2(a+b+c)\le 10+abc\ }$ . Indeed, can suppose w.l.o.g. $a\le b\le c$ . Thus,

$\underline{2(a+b+c)}=\left[2\cdot b+2\cdot c+(-1)\cdot a\right]+3a\stackrel{(C.B.S)}{\le}$ $\sqrt {\left[2^2+2^2+(-1)^2\right]\cdot\left[b^2+c^2+a^2\right]}+3a=9+3a\stackrel{(*)}{\le}\underline{10+abc}$ .


Proposed problem 2. Let $\left\{\begin{array}{c} a>0\ ,\ b>0\\\ (a-1)(b-1)>0\end{array}\right\|\ \implies\ \boxed{\ a^b + b^a > 1 + ab + (1-a)(1-b)\cdot\min \{1,ab\}\ }$ (CRUX, 3260/4/2007).

Proof. I"ll use the well-known Bernoulli's inequality : $\left\{\begin{array}{cccc} 0<a<1 & \Longrightarrow & \left( \forall \right) x>-1 \ , & (1+x)^a\le 1+ax\\\\ 1\le a & \Longrightarrow & \left( \forall \right) x>-1 \ , & (1+x)^a\ge 1+ax\end{array}\right\|$ .

The first case. $0<a,b\le 1\Longrightarrow \min \{ 1,ab\}=ab$ . Thus, $\left(\frac 1a\right)^b=$ $\left[ 1+\left( \frac 1a -1\right)\right]^b\le $ $\frac{a+b-ab}{a}$ $\Longrightarrow $ $a^b\ge \frac{a}{a+b-ab}$ .

Analogously, $b^a\ge \frac{b}{a+b-ab}$. Thus, $a^b+b^a\ge $ $\frac{a+b}{a+b-ab}=$ $1+ab+\frac{ab(1-a)(1-b)}{a+b-ab}$ . Since $a+b-ab=a+b(1-a)>0$ ,

$\frac{ab}{a+b-ab}\ge ab$ $\Longleftrightarrow$ $ (1-a)(1-b)\ge 0$ , what is truly. Thus, $\frac{ab(1-a)(1-b)}{a+b-ab}\ge$ $ ab(1-a)(1-b)$ .

Therefore, $a^b+b^a\ge $ $1+ab+ab(1-a)(1-b)=$ $1+ab+(1-a)(1-b)\cdot \min \{1,ab\}$ .

The second case. $1\le a,\ 1\le b $ $\Longrightarrow $ $\min \{1,ab\}=1$ . Thus, $a^b\ge 1+b(a-1)$ , i.e. $a^b\ge 1+ab-b$ .

Analogously, $b^a\ge 1+ab-a$ . Thus, $a^b+b^a\ge $ $1+ab+(1-a)(1-b)=$ $1+ab+(1-a)(1-b)\cdot \min \{1,ab\}$ . See here.
This post has been edited 49 times. Last edited by Virgil Nicula, Nov 22, 2015, 6:30 PM

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29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by OlympusHero, Dec 30, 2020, 6:08 PM

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

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  • Total entries: 456
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