Easy inequality

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Easy inequality

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#1 h Dec 22, 2012, 4:03 PM • 2 Y

Y by Adventure10, Mango247

For $a+b+c=1$ and $a,b,c > 0$
Prove that: $\frac{13}{27} \le a^2+b^2+c^2+4abc \le \frac{1}{2}$

This post has been edited 1 time. Last edited by hieu1411997, Dec 23, 2012, 2:29 AM

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arqady

#2 h Dec 22, 2012, 10:09 PM • 2 Y

Y by Adventure10, Mango247

hieu1411997 wrote:

For $a+b+c=1$ and $a,b,c \ge 0$
Prove that: $\frac{13}{27} \le a^2+b^2+c^2+4abc \le \frac{1}{2}$

The right inequality is wrong. Try $a=1$ and $b=c=0$ .

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#3 h Dec 23, 2012, 2:29 AM • 1 Y

Y by Adventure10

arqady wrote:

hieu1411997 wrote:

For $a+b+c=1$ and $a,b,c \ge 0$
Prove that: $\frac{13}{27} \le a^2+b^2+c^2+4abc \le \frac{1}{2}$

The right inequality is wrong. Try $a=1$ and $b=c=0$ .

Sorry, $a,b,c>0$

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Sayan

#4 h Dec 23, 2012, 2:55 AM • 1 Y

Y by Adventure10

Then try $a=0.8, b=0.1, c=0.1$ for the right side

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sqing

#5 h Jan 31, 2020, 8:43 AM • 1 Y

Y by Adventure10

hieu1411997 wrote:

For $a+b+c=1$ and $a,b,c > 0$
Prove that: $\frac{13}{27} \le a^2+b^2+c^2+4abc \le \frac{1}{2}$

Let $a,b,c$ be side lengths of a triangle with $a+b+c = 1$ . Prove that $a^2 +b^2 +c^2 +4abc \le \frac12$ 1990 ITAMO
Let $a, b$ and $c$ be the lengths of the sides of a triangle $ABC$ . If $a+b+c = 1.$ Prove that

${{1}\over{a^2}}+{{1}\over{b^2}}+{{1}\over{c^2}}+{{1}\over{4abc}}\geq \ 32$

This post has been edited 1 time. Last edited by sqing, Jan 31, 2020, 8:47 AM

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