Hard inequality

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Source: MEMO 2012

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49 posts

soad

#1 h Sep 13, 2012, 3:01 PM • 1 Y

Y by Adventure10

Hello. How to prove this hard inequality ?

Let $a,b,c$ be a positive real numbers that satisfy $abc=1$ . Prove that:

$\sqrt{9+16a^2}+\sqrt{9+16b^2}+\sqrt{9+16c^2} \ge 3 + 4(a+b+c)$ .

This problem was not solved by any team.

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1009 posts

#2 h Sep 13, 2012, 4:11 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

soad wrote:

Hello. How to prove this hard inequality ?

Let $a,b,c$ be a positive real numbers that satisfy $abc=1$ . Prove that:

$\sqrt{9+16a^2}+\sqrt{9+16b^2}+\sqrt{9+16c^2} \ge 3 + 4(a+b+c)$ .

This problem was not solved by any team.

Actually, I solved it at that competition, but after I came out. While I was solving it during the competition, because of my mistake in calculating derivative, I wrote that it's enough to prove inequality when $a=b$ because of EV theorem, but that argument follows from HCF theorem, so they gave us just 2 points on that problem.

Anyway, this is my proof:

Let's define $x=\ln a, \; y=\ln b, z= \ln c$ . Now we have $x+y+z=0$ , and inequality becomes:

$\sum \left( \sqrt{9+16e^{2x}} - 4e^x \right) \ge 3$

Now, we define $f(u)=\sqrt{9+16e^{2u}} - 4e^u$ and it's convex for $u\ge 0$ . Hence, because of HCF theorem we need to prove it just when $a=b$ . That is, we need to prove: If $a^2c=1$ , then:

$2 \sqrt{9+16a^2}+\sqrt{9+16c^2} \ge 3+4(2a+c)$

After squaring both sides, it remains to prove:

$\sqrt{(9+16a^2)(9+16c^2)} \ge 16ac+12a+6c-9$

What is after squaring and using $c=\frac{1}{a^2}$ equivalent to:

$\frac{12}{a^4}(a-1)^2(18a^3+4a^2+2a+9) \ge 0$

What is obviously true.

Equality holds iff $a=b=c=1$ . $\blacksquare$

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#3 h Feb 4, 2015, 6:53 AM • 2 Y

Y by Adventure10, Mango247

Assume $(b-1)(c-1)\geqslant 0 \Leftrightarrow bc+1\geqslant b+c$ (Dirichlet)
Using Minkovsky's inequality: $\sqrt{9+16b^2}+\sqrt{9+16c^2}\geqslant \sqrt{36+16(b+c)^2}$
We have:
$\sqrt{36+16(b+c)^2}-4(b+c)\geqslant\sqrt{36+16\left(\dfrac{1}{a}+1\right)^2}-\dfrac{4}{a}-4$
So we need prove: $\sqrt{9+16a^2}+\sqrt{36+16\left(\dfrac{1}{a}+1\right)^2}-\dfrac{4}{a}-4\geqslant 4a+3 \Leftrightarrow a^2(a-1)^2 \geqslant 0$ (True)
Equality occurs when and only when $a=b=c=1$

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#4 h Mar 1, 2017, 3:44 PM • 2 Y

Y by Adventure10, Mango247

http://epsilon.ro/wp-content/uploads/2012/09/2013_Matematica_Internationala_Concursul-MEMO2012_Solutii.pdf

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