Inequality with three variables

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Inequality with three variables

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Source: Turkey JBMO Team Selection Test 2013, P4

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1345 posts

#1 h May 31, 2013, 7:29 PM • 4 Y

Y by sdsert, microsoft_office_word, Adventure10, cubres

For all positive real numbers $a, b, c$ satisfying $a+b+c=1$ , prove that

$\frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 1- ab-bc-ca$

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30203 posts

arqady

#2 h May 31, 2013, 8:02 PM • 3 Y

Y by Adventure10, ehuseyinyigit, cubres

After using Cauchy-Schwarz we'll get something obvious.
Also SOS helps here.
This inequality is very easy.

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1345 posts

#3 h May 31, 2013, 8:11 PM • 3 Y

Y by Adventure10, Mango247, cubres

arqady wrote:

After using Cauchy-Schwarz we'll get something obvious.
Also SOS helps here.
This inequality is very easy.

Yes, it is obvious

.

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15142 posts

#4 h May 31, 2013, 8:32 PM • 4 Y

Y by Adventure10, Mango247, Nurefsan17, cubres

$\sum \frac{a^4+5b^4}{a(a+2b)} =\sum \frac{a^4}{a(a+2b)} + 5\sum \frac{b^4}{a(a+2b)} \geq \frac {(\sum a^2)^2} {\sum a(a+2b)} + 5\frac {(\sum a^2)^2} {\sum a(a+2b)}$ .
Now, $\sum a(a+2b) = \sum a^2 + 2\sum ab = (\sum a)^2 = 1$ , and $3\sum a^2 \geq (\sum a)^2 = 1$ , so the above continues $= 6(\sum a^2)^2 \geq 2\sum a^2$ . Write $2\sum a^2 \geq 1 - \sum ab$ as $4\sum a^2 + 2\sum ab \geq 2$ , and indeed, $4\sum a^2 + 2\sum ab = 3\sum a^2 + (\sum a)^2 \geq 1+1=2$ .

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#5 h Jun 1, 2013, 9:21 PM • 6 Y

Y by aliyev95uz, ZRH, ywq233, Adventure10, Mango247, cubres

For all $t> 0$ one has $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ because this is (for $t>0$ !) equivalent to $(t-1)^2(4t+15) \geq 0$ . For $t = a/b$ , $t = b/c$ and $t = c/a$ this implies that $\begin{aligned}\sum \frac{a^4+5b^4}{a(a+2b)} & \geq \sum \frac{1}{3}(3a^2-10ab+13b^2) \\ & = \frac{1}{3}\left(16(a^2+b^2+c^2) - 10(ab+bc+ca)\right).\end{aligned}$ So it suffices to show that $16(a^2+b^2+c^2) \geq 3 + 7(ab+bc+ca)$ . Using $a+b+c = 1$ , we write $3 = 3(a+b+c)^2$ to get the equivalent $a^2+b^2+c^2\geq ab+bc+ca$ , which is true by $\sum (a-b)^2\geq 0$ .

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sqing

#6 h Jun 2, 2013, 1:58 AM • 3 Y

Y by Adventure10, Mango247, cubres

For all positive real numbers $a, b, c$ , prove that $\frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 2(a^2+b^2+c^2).$

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3212 posts

#7 h Jun 2, 2013, 3:50 AM • 2 Y

Y by Adventure10, cubres

Kurt Gödel wrote:

For all $t> 0$ one has $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ because this is (for $t>0$ !) equivalent to $(t-1)^2(4t+15) \geq 0$ . For $t = a/b$ , $t = b/c$ and $t = c/a$ this implies that $\begin{aligned}\sum \frac{a^4+5b^4}{a(a+2b)} & \geq \sum \frac{1}{3}(3a^2-10ab+13b^2) \\ & = \frac{1}{3}\left(16(a^2+b^2+c^2) - 10(ab+bc+ca)\right).\end{aligned}$ So it suffices to show that $16(a^2+b^2+c^2) \geq 3 + 7(ab+bc+ca)$ . Using $a+b+c = 1$ , we write $3 = 3(a+b+c)^2$ to get the equivalent $a^2+b^2+c^2\geq ab+bc+ca$ , which is true by $\sum (a-b)^2\geq 0$ .

How could you find this inequality $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ ??

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1093 posts

#8 h Jun 2, 2013, 6:28 AM • 4 Y

Y by sqing, Adventure10, Mango247, cubres

sqing wrote:

For all positive real numbers $a, b, c$ , prove that $\frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 2(a^2+b^2+c^2).$

___________________
We have:
$\frac{a^4+5b^4}{a(a+2b)}+\frac{b^4+5c^4}{b(b+2c)}+\frac{c^4+5a^4}{c(c+2a)}\geq 2(a^2+b^2+c^2).$
1)
$\Leftrightarrow\sum\frac{(a^2)^2}{a(a+2b)}+5\sum\frac{(b^2)^2}{b(b+2c)}\geq 2(a^2+b^2+c^2).$
We apply the Cauchy-Schwarz inequality, we get:
$LHS\ge\frac{6(a^2+b^2+c^2)^2}{(a+b+c)^2}=\frac{2(a^2+b^2+c^2)3(a^2+b^2+c^2)}{(a+b+c)^2} \ge 2(a^2+b^2+c^2)$ since $3(a^2+b^2+c^2)\ge (a+b+c)^2$
The proof is ended!
_________
Sandu Marin

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#9 h Jun 2, 2013, 7:38 AM • 2 Y

Y by Adventure10, cubres

Kurt Gödel wrote:

$\begin{aligned}\sum \frac{a^4+5b^4}{a(a+2b)} & \geq \frac{1}{3}\left(16(a^2+b^2+c^2) - 10(ab+bc+ca)\right).\end{aligned}$

It is true the following
$\frac{a^4+5b^4}{a(a+2b)}+\frac{b^4+5c^4}{b(b+2c)}+\frac{c^4+5a^4}{c(c+2a)}\geq \frac{31}{3}(a^2+b^2+c^2)-\frac{25}{3}(ab+bc+ca).$

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mathuz

#10 h Jun 2, 2013, 7:53 AM • 3 Y

Y by Adventure10, Mango247, cubres

using by AM-GM inequality: ...

The equality holds: $a=b=c=\frac{1}{3}.$

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#11 h Jun 2, 2013, 10:33 AM • 6 Y

Y by shinichiman, ArefS, ywq233, Adventure10, Mango247, cubres

sqing wrote:

For all positive real numbers $a, b, c$ , prove that $\frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 2(a^2+b^2+c^2).$

By my proof it suffices to show that $\frac{1}{3}\left(16(a^2+b^2+c^2)-10(ab+bc+ca)\right)\geq 2(a^2+b^2+c^2)$ which is obviously true.

shinichiman wrote:

Kurt Gödel wrote:

For all $t> 0$ one has $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ because this is (for $t>0$ !) equivalent to $(t-1)^2(4t+15) \geq 0$ . For $t = a/b$ , $t = b/c$ and $t = c/a$ this implies that $\begin{aligned}\sum \frac{a^4+5b^4}{a(a+2b)} & \geq \sum \frac{1}{3}(3a^2-10ab+13b^2) \\ & = \frac{1}{3}\left(16(a^2+b^2+c^2) - 10(ab+bc+ca)\right).\end{aligned}$ So it suffices to show that $16(a^2+b^2+c^2) \geq 3 + 7(ab+bc+ca)$ . Using $a+b+c = 1$ , we write $3 = 3(a+b+c)^2$ to get the equivalent $a^2+b^2+c^2\geq ab+bc+ca$ , which is true by $\sum (a-b)^2\geq 0$ .

How could you find this inequality $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ ??

The right hand side of the original inequality is quadratic in disguise, so we want to bound each term $(a^4+5b^4)/(a(a+2b))$ by something quadratic ("isolated fudging"). By dehomogenizing with respect to $b = 1$ , this means that you want to find $\alpha, \beta, \gamma$ such that $t^4 + 5 \geq t(t+2)(\alpha t^2 + \beta t + \gamma)$ with equality in $t = 1$ . This means that the polynomial $f(t) = t^4+5 - t(t+2)(\alpha t^2+\beta t + \gamma)$ should have a double zero in $t = 1$ : this already gives two equations for $\alpha,\beta,\gamma$ . The third equation comes from the quadratic quotient $f(t)/(t-1)^2$ : it should have a nonnegative leading coefficient and only negative roots (or none). Different people handle the quadratic term differently, so I guess that's why MariusStanean got another bound. He actually got a better bound, and I think he did this by taking $\frac{1}{2}g''(1)(t-1)^2 + g'(1)(t-1) + g(1)$ as bound, where $g(t) = (t^4+5)/t(t+2)$ .

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#12 h Jun 2, 2013, 2:27 PM • 2 Y

Y by Adventure10, cubres

Thank you very much!!!

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#13 h Oct 21, 2016, 7:28 PM • 4 Y

Y by Adventure10, Mango247, wizixez, cubres

Really, isn't it too easy, especially for Turkey? It can be solved in a minute...

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168 posts

#14 h Jan 1, 2025, 12:58 PM • 1 Y

Y by cubres

@above really.1 min solve by T2

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#15 h Apr 5, 2025, 4:55 PM • 1 Y

Y by cubres

quite a good ineq to teach young students about cauchy or titu

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