Inequalities

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#1 h Oct 26, 2017, 3:31 PM • 1 Y

Y by Adventure10

1, Let $a,b,c$ be positive numbers such that $a^2+b^2=c^2$ . Find the minimum value of:
$P=\frac{a^2(b+c)+b^2(a+c)}{abc}$ 2, Let $a,b,c$ be positive numbers such that $abc=1$ . Find the maximum value of:
$P=\sqrt{\frac{2}{1+a}}+\sqrt{\frac{2}{1+b}}+\sqrt{\frac{2}{1+c}}$

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S_SRW

#2 h Oct 26, 2017, 5:14 PM • 2 Y

Y by Adventure10, Mango247

In the first question, we can substitute a and b to $csinx$ and $ccosx$ respectively.And then we verify $P$ in term of $sinx$ and $cosx$ . $P=(sinx+cosx)+\frac{2}{(sinx+cosx)^2-1}$ define $f(y)=y+\frac{2}{y^2-1}$ on $y \in [0,\sqrt{2}]$
Thus, we can find local minimum on this function.

This post has been edited 3 times. Last edited by S_SRW, Oct 26, 2017, 5:41 PM

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#5 h Oct 27, 2017, 11:13 AM • 2 Y

Y by Adventure10, Mango247

Does anyone have simplier solutions?

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1900 posts

#6 h Oct 27, 2017, 1:19 PM • 2 Y

Y by Adventure10, Mango247

I have something like this:
$P=\frac{a^2(b+c)+b^2(a+c)}{abc}=\frac{ab(a+b)+c(a^2+b^2)}{abc}=\frac{ab(a+b)+c^3}{abc}=\frac{a}{c}+\frac{b}{c}+\frac{c^2}{ab}$
By inequality between arithmetic mean and geometric mean we have:
$\frac{\frac{a}{c}+\frac{b}{c}+\frac{c^2}{ab}}{3} \ge \sqrt[3]{\frac{a}{c}\cdot\frac{b}{c}\cdot\frac{c^2}{ab}}=1$
The equality holds iff $\frac{a}{c}=\frac{b}{c}=\frac{c^2}{ab} \iff a=b=c$ . Putting this into $a^2+b^2=c^2$ you get $c=0$ which is a contradiction. So
$P>3$

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1900 posts

#7 h Oct 27, 2017, 1:32 PM • 2 Y

Y by Adventure10, Mango247

Finishing S_SRW idea.
With given assumptions $a, b, c$ are the sides of rectangular triangle.
$sinx+cosx=\frac{a+b}{c}>1$ which holds by triangle inequality.
$\frac{sinx+cosx}{2}\le \sqrt{\frac{sin^2x+cos^2x}{2}}=\sqrt{\frac{1}{2}}$
So $1<(sinx+cosx)^2\le 2 \implies 0<(sinx+cosx)^2-1\le 1\implies \frac{2}{(sinx+cosx)^2-1}\ge 2$
That's why $P=(sinx+cosx)+\frac{2}{(sinx+cosx)^2-1}>3$

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727 posts

#8 h Oct 27, 2017, 1:44 PM • 2 Y

Y by Adventure10, Mango247

From your idea:

WolfusA wrote:

I have something like this:
$P=\frac{a^2(b+c)+b^2(a+c)}{abc}=\frac{ab(a+b)+c(a^2+b^2)}{abc}=\frac{ab(a+b)+c^3}{abc}=\frac{a}{c}+\frac{b}{c}+\frac{c^2}{ab}$
By inequality between arithmetic mean and geometric mean we have:

$\frac{a}{c}+\frac{b}{c}+\frac{c^2}{ab} \ge 3\sqrt[3]{\frac{a}{c}\cdot\frac{b}{c}\cdot\frac{c^2}{2\sqrt{2}ab}}+\frac{(4-\sqrt{2})c^2}{4ab} \ge 3\sqrt[3]{\frac{1}{2\sqrt{2}}}+\frac{(4-\sqrt{2})2ab}{4ab}=2+\sqrt{2}$
The equality holds if $c=\sqrt{2}a=\sqrt{2}b$ .

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#9 h Oct 28, 2017, 1:29 PM • 2 Y

Y by Adventure10, Mango247

Any ideas for the second inequality?

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1900 posts

#10 h Oct 29, 2017, 12:10 PM • 2 Y

Y by Adventure10, Mango247

It suffices to prove that when $a, b, c>0$ and $abc=1$ then $\frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1}\ge 0$ . I'm actually trying to prove it.

This post has been edited 1 time. Last edited by WolfusA, Oct 29, 2017, 12:11 PM
Reason: typo

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1900 posts

#11 h Oct 29, 2017, 12:44 PM • 2 Y

Y by Adventure10, Mango247

It's easy to find out (multiplying everything and using $2abc=2$ ) that $\frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1}=\frac{(a-1)(b-1)(c-1)}{(a+1)(b+1)(c+1)}$ . Suppose, that $min\lbrace{a,b,c\rbrace}=c \wedge max\lbrace{a,b,c\rbrace}=a$ . Then $a-1\ge0 \wedge c-1\le 0 \implies (a-1)(c-1)\le0$ . So $b\le 1 \implies (a-1)(b-1)(c-1)\ge0$ and $b>1 \implies (a-1)(b-1)(c-1)\le0$ . When $b\le 1$ the prove will be finished, but now we have to figure out what if $b> 1$ ?

This post has been edited 3 times. Last edited by WolfusA, Oct 29, 2017, 12:47 PM

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#12 h Oct 29, 2017, 1:07 PM • 1 Y

Y by Adventure10

The function $f(x)= \sqrt{\frac{2}{1+x}}$ is decreasing for $x\ge -1$ , so if $a\ge b>1\ge c>0$ , then $P\le \sqrt{\frac{2}{1+1}}+\sqrt{\frac{2}{1+1}}+\sqrt{\frac{2}{1+0}}<2+\sqrt2$

Assume as above, that $min\lbrace{a,b,c\rbrace}=c \wedge max\lbrace{a,b,c\rbrace}=a$ .
If $b>1$ , then read two first sentences from this post.
If $b\le 1$ , then as in post above $b\le 1 \implies (a-1)(b-1)(c-1)\ge0$ , so $\frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1}=\frac{(a-1)(b-1)(c-1)}{(a+1)(b+1)(c+1)}\ge 0$ . Adding to both sides $\frac{2}{a+1}+\frac{2}{b+1}+\frac{2}{c+1}$ gives $3\ge \frac{2}{a+1}+\frac{2}{b+1}+\frac{2}{c+1}$ .
Now use the inequality between arithmetic and quadratic mean:
$\frac{\sqrt{\frac{2}{1+a}}+\sqrt{\frac{2}{1+b}}+\sqrt{\frac{2}{1+c}}}{3}\le \sqrt{\frac{\frac{2}{a+1}+\frac{2}{b+1}+\frac{2}{c+1}}{3}} \le \sqrt{\frac{3}{3}}=1$ . So here $P\le 3$ with equality iff $a=b=c=1$ .

This post has been edited 1 time. Last edited by WolfusA, Oct 29, 2017, 1:07 PM

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#13 h Dec 17, 2017, 4:06 AM • 1 Y

Y by Adventure10

you can use chebyshev to solve problem 2 :v

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#15 h Jan 4, 2018, 5:01 AM • 1 Y

Y by Adventure10

2 Let $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ where $x,y,z$ are positive, using Cauchy-Schwarz, we have:
$P=\sqrt{\frac{2}{1+a}}+\sqrt{\frac{2}{1+b}}+\sqrt{\frac{2}{1+c}}=\sqrt{\frac{2y(x+z)}{(y+x)(x+z)}}+\sqrt{\frac{2z(x+y)}{(z+y)(x+y)}}+\sqrt{\frac{2x(y+z)}{(x+z)(y+z)}} \le \sqrt{(4xy+4yz+4zx)\frac{2x+2y+2z}{(x+y)(y+z)(z+x)}}\le 3$

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#16 h Feb 25, 2018, 6:16 PM • 2 Y

Y by Adventure10, Mango247

Task 2 is equivalent to China West Math Olympiad 2004
https://artofproblemsolving.com/community/c6h17549_nice_looking_but_hard_inequality

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#17 h Feb 25, 2018, 8:30 PM • 2 Y

Y by Adventure10, Mango247

If we delete the fact $a^2+b^2=c^2$ , but suppose that $a<b+c, b<a+c, c<a+b$ , then $P>3$ stays true?

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#18 h Feb 26, 2018, 6:03 PM • 2 Y

Y by Adventure10, Mango247

I think there would be a counterexample, but certainly for all $a,b,c>0$ we have $P>2$

This post has been edited 1 time. Last edited by WolfusA, Feb 26, 2018, 6:03 PM

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#19 h Mar 2, 2018, 3:44 AM • 2 Y

Y by Adventure10, Mango247

WolfusA wrote:

It's easy to find out (multiplying everything and using $2abc=2$ ) that $\frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1}=\frac{(a-1)(b-1)(c-1)}{(a+1)(b+1)(c+1)}$ . Suppose, that $min\lbrace{a,b,c\rbrace}=c \wedge max\lbrace{a,b,c\rbrace}=a$ . Then $a-1\ge0 \wedge c-1\le 0 \implies (a-1)(c-1)\le0$ . So $b\le 1 \implies (a-1)(b-1)(c-1)\ge0$ and $b>1 \implies (a-1)(b-1)(c-1)\le0$ . When $b\le 1$ the prove will be finished, but now we have to figure out what if $b> 1$ ?

Dear wolf usa
are you sure

$\frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1}=\frac{(a-1)(b-1)(c-1)}{(a+1)(b+1)(c+1)}$

it is checked they are with different sign.

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1900 posts

#20 h Mar 2, 2018, 2:34 PM • 1 Y

Y by Adventure10

It should be $\frac{a-1}{a+1}+\frac{b-1}{b+1}+\frac{c-1}{c+1}=-\frac{(a-1)(b-1)(c-1)}{(a+1)(b+1)(c+1)}$

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CHINO

#21 h Mar 10, 2018, 4:13 PM • 2 Y

Y by Adventure10, Mango247

1) a,b,c>0, a^2+b^2=c^2, Calcule el minimo de P=[a^2(b+c)+b^2(a+c)]/abc
Sea a=csent, b=ccost
P=sent+1/(sentcost)+cost
P>=3 luego el mínimo de P es 3
2) a,b,c>0, abc=1, Calcule el maximo valor de S=sqrt[2/(1+a)]+sqrt[2/(1+b)]+sqrt[2/(1+c)]
MA<=MC
S=sqrt(2/1+a)+sqrt(2/1+b)+sqrt(2/1+c)<=3sqrt([2/(1+a)+2/(1+b)+2/(1+c)]/3).......(I)
pero (1-a)/(1+a)+(1-b)/(1+b)+(1-c)/(1+c)>=0
2/(1+a)+2/(1+b)+2/(1+c)<=3.............................................................................(II)
De (I) y (II)
S<=3 luego el minimo de S es 3

This post has been edited 1 time. Last edited by CHINO, Mar 11, 2018, 2:17 PM
Reason: s

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sqing

#23 h May 21, 2020, 12:13 AM • 2 Y

Y by Mango247, Mango247

Let $a,b,c$ be positive numbers such that $a^2=b^2+c^2$ . Prove that $\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 5+3\sqrt{2}$ *

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