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328 posts

#1 h Aug 13, 2012, 5:11 PM • 2 Y

Y by Adventure10, Mango247

Let a,b,c be positive real. Prove that:
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}-3 \ge 3 \Big(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\Big)$

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1345 posts

#2 h Aug 13, 2012, 7:22 PM • 2 Y

Y by teomihai, Adventure10

The problem is equivalent to the following one:
$x^2+y^2+z^2 +6 \geq 3(x+y+z)$
where $x, y, z$ are positive real numbers satisfying $xyz=1$ .

Let $x+y+z=u$ and $xy+yz+zx=v$ . Then we have $u^3+9 \geq 4uv$ by Schur's inequality.

Our inequality to be proved is $u^2 -3u+6 \geq 2v$ hence it is enough to show that
$2u(u^2-3u+6) \geq u^3+9$
which is equivalent to
$(u-3)[u(u-3)+3] \geq 0$
which is true since $u \geq 3$ by AM-GM.

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#3 h Aug 13, 2012, 11:35 PM • 1 Y

Y by Adventure10

(1)Let a,b,c be positive real. Prove that:
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}-3 \ge {(9-3\sqrt{3}})\Big(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\Big)$
(2)Let a,b,c,d be positive real. Prove that:
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{d^2}+\frac{d^2}{a^2}-4 \ge {\frac{7}{2}\Big(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}-4\Big)}$
Remark

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30154 posts

arqady

#4 h Aug 14, 2012, 6:19 PM • 2 Y

Y by Adventure10, Mango247

Lastnightstar wrote:

(1)Let a,b,c be positive real. Prove that:
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}-3 \ge {(9-3\sqrt{3}})\Big(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\Big)$

We need to prove that ${x^2+y^2+z^2-3\geq(9-3\sqrt{3}})(x+y+z-3)$ ,
where $x$ , $y$ and $z$ are positives such that $xyz=1$ .
Let $x+y+z=3u$ , $xy+xz+yz=3v^2$ and $xyz=w^3$ .
Hence, our inequality is equivalent to $f(v^2)\geq0$ , where $f$ is a decreasing linear function.
Hence, $f$ gets a minimal value, when $v^2$ gets a maximal value, which happens,
when two numbers from $\{x,y,z\}$ are equal.
Thus, it remains to prove the last inequality for $y=z$ and $x=\frac{1}{y^2}$ , which gives
$(y-1)^2(2y-1-\sqrt3)^2(y^2+2(2\sqrt3-3)y+2-\sqrt3)\geq0$ , which is true.

Let $a$ , $b$ and $c$ are positive numbers. Prove that:
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}-3 \ge 2\Big(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}-3\Big)$

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1345 posts

#5 h Aug 14, 2012, 6:39 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

Let $a$ , $b$ and $c$ are positive numbers. Prove that:
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}-3 \ge 2\Big(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}-3\Big)$

Schur kills it immediately.

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328 posts

#6 h Aug 15, 2012, 3:11 AM • 2 Y

Y by Adventure10, Mango247

A generalization of it:
1. Let x,y,z be positive real which satisfy: $xyz=1$ .
Prove that: $x^3+y^3+z^3-3 \ge 8(x+y+z-3)$
p/s: I think it is not hard.
2. (Test for VMO) Let x,y,z be positive real which satisfy: $xyz=1$
Prove that: $x^3+y^3+z^3+6 \ge (x+y+z)^2$

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30154 posts

arqady

#7 h Aug 15, 2012, 4:00 AM • 2 Y

Y by Adventure10, Mango247

The both inequalities we can prove by $uvw$ .

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328 posts

#8 h Aug 15, 2012, 4:10 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

The both inequalities we can prove by $uvw$ .

Yes, I think you can solve it by this method.
(this method is very known, as you)

Other genelization of it:
Given integer $n \ge 2$ . Let $x_1,x_2,...,x_n$ be positive real which satisfy: $x_1x_2...x_n=1$
Prove that: $x_1^2+x_2^2+...+x_n^2-n \ge \Big( 2+\frac{1}{n}\Big) \Big(x_1+x_2+...+x_n-n \Big)$

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559 posts

#9 h Aug 15, 2012, 4:28 AM • 2 Y

Y by Adventure10, Mango247

perfect_square wrote:

Given integer $n \ge 2$ . Let $x_1,x_2,...,x_n$ be positive real which satisfy: $x_1x_2...x_n=1$
Prove that: $x_1^2+x_2^2+...+x_n^2-n \ge \Big( 2+\frac{1}{n}\Big) \Big(x_1+x_2+...+x_n-n \Big)$

And we have
$x_1^{n-1}+x_2^{n-1}+...+x_n^{n-1}-n \ge 2(x_1^{-1}+x_2^{-1}+...+x_n^{-1}-n)(n\ge3)$ See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=492196

Let $x_1,x_2,...,x_n$ be positive real which satisfy: $x_1x_2...x_n=1,$
find the best $k(n)$ such that $x_1^2+x_2^2+...+x_n^2-n \ge k (n)\Big(x_1+x_2+...+x_n-n \Big)$

We have $k(2)=4,k(3)=3.80384757...(k^2-18k+54=0),$
$k(4)=3.56190850...(3k^5-124k^4+1600k^3-5376k^2+5120k-4096=0).$

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41255 posts

sqing

#10 h Sep 2, 2019, 12:11 AM • 1 Y

Y by Adventure10

perfect_square wrote:

Let a,b,c be positive real. Prove that:
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}-3 \ge 3 \Big(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\Big)$

Let $a,b,c$ be positive reals. Prove that
$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} \ge 3 \Big(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-2\Big)\ge \frac{3(a+b)(b+c)(c+a)}{8abc},$ $\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3} \ge \frac{a^3+b^3+c^3}{3abc}.$

Attachments:

This post has been edited 3 times. Last edited by sqing, Sep 2, 2019, 11:46 PM

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sqing

#11 h Sep 1, 2021, 2:09 PM

Y by

crazyfehmy wrote:

$x^2+y^2+z^2 +6 \geq 3(x+y+z)$ where $x, y, z$ are positive real numbers satisfying $xyz=1$ .

h
SXTB,(8)2021,Q2620
Let $a, b, c,$ be non-zero real numbers such $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Prove that $a^2 + b^2 + c^2 + 6 \ge 3 (a+b+c) .$ Let $a,b,c$ be real numbers such that $ab+bc+ca=1.$ Prove that $a^2+b^2+c^2+11\geq 6(a+b+c).$ If $a+b+c \geq 2,$ then $a^2+b^2+c^2+4 \geq 3(a+b+c),$ h

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#12 h Sep 4, 2021, 12:03 PM

Y by

crazyfehmy wrote:

The problem is equivalent to the following one:
$x^2+y^2+z^2 +6 \geq 3(x+y+z)$ where $x, y, z$ are positive real numbers satisfying $xyz=1$ .

After you obtain this inequality it suffice to apply Chebyshev and AM-GM like this :
$x(x-3) + y(y-3)+ z(z-3) +6 \geq$ $(xyz)^{\frac{1}{3}} . (3(xyz)^{\frac{1}{3}} -9) +6 = 0 .$

This post has been edited 3 times. Last edited by edemafa, Sep 4, 2021, 12:05 PM

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