Please Help! I'm Getting Stuck in Inequality

This inequality problem is from Phan Boi Chau High School for the Gifted Grade 10 Selection Test, Nghe An, Vietnam (Yes I'm very bad at inequalities):

Let $a,b,c$ be positive reals. Find the minimum value of $P=\dfrac{a^2}{(a+b)^2}+\dfrac{b^2}{(b+c)^2}+\dfrac{c}{4a}$ .

I've done $\dfrac{c}{4a}=\dfrac{c^2}{4ac}\geq\dfrac{c^2}{(c+a)^2}$ .

I think the minimum value is $\dfrac34$ , so the problem is to prove $\displaystyle \sum_\text{cyc}\dfrac{a^2}{(a+b)^2}\geq\dfrac34$ . But I still can't.

Can you help?

#Algebra

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

After setting, $x = \frac{b}{a}, y = \frac{c}{b}, z = \frac{a}{c}$ this becomes:
For $x, y, z \in R_+$ such that $xyz = 1$ , $\left(\frac{1}{1 + x}\right)^2 + \left(\frac{1}{1 + y}\right)^2 + \left(\frac{1}{1 + z}\right)^2 \geq \frac{3}{4}$

I believe Lagrange Multipliers is the best way here. I'm trying to find a non-calculus approach, and I've posted the question on AoPS as well.

Ameya Daigavane - 5 years ago

This Inequality was used in this problem,Viet Nam TST 2006 inequality problem: PROBLEM

Son Nguyen - 4 years, 11 months ago

Continue the solution of @Ameya Daigavane . Using Cauchy-Schwarz $\left(\frac{1}{1 + x}\right)^2 + \left(\frac{1}{1 + y}\right)^2 + \left(\frac{1}{1 + z}\right)^2 \geq\frac{1}{3}\bigg(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\bigg)^2$ So now we need to have $f(x,y,z)=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\geq\frac{3}{2}$ . Without losing generality, we assume that $x\geq y\geq z$ , then $f(x,y,z)\geq f(x,y,y)$ .

Now we consider $f(x,y,y)$ . Since $x\geq y$ and $xy^2=1$ we can have $y\leq 1$ and $x=\frac{1}{y^2}$ . $f(x,y,y)=\frac{1}{x+1}+\frac{2}{y+1}=\frac{y^2}{y^2+1}+\frac{2}{y+1}=\frac{y^3+3y^2+2}{(y^2+1)(y+1)}$ Now we prove $f(x,y,y)\geq\frac{3}{2}$ $\frac{y^3+3y^2+2}{(y^2+1)(y+1)}\geq\frac{3}{2}$ $\Leftrightarrow\frac{2y^3+6y^2+4-3(y^2+1)(y+1)}{2(y+1)(y^2+1)}\geq 0$ $\Leftrightarrow\frac{(1-y)^3}{2(y+1)(y^2+1)}\geq 0 \ (True \ since \ y\leq 1)$ So $f(x,y,y)\geq\frac{3}{2}\Rightarrow f(x,y,z)\geq\frac{3}{2}$ , therefore $P\geq\frac{3}{4}$ . The equality holds when $x=y=z=1$ or $a=b=c$ . @Tran Quoc Dat , you can check out this solution, I don't know whether it's ok for grade 9 mathematically specialized students to apply this method

P C - 4 years, 12 months ago

Is this "mixing of variables"? I think there's some calculus theory behind this method.
Any non calculus approach?

Ameya Daigavane - 4 years, 12 months ago

I've tried but looks like there's no non calculus solution to this

P C - 4 years, 12 months ago

Did you mean the $f(x, y, z) \geq f(x, y, y)$ ?

A Former Brilliant Member - 4 years, 12 months ago

@A Former Brilliant Member – Oh no, he changed it, it was $f(x,y,z) \geq f(x, \sqrt{yz}, \sqrt{yz})$ I think.

Ameya Daigavane - 4 years, 11 months ago

@Ameya Daigavane – I've to change because later on, I found out that actually, $f(x,y,z)\leq f(x,\sqrt{yz},\sqrt{yz})$

P C - 4 years, 11 months ago

@Ameya Daigavane – Ok. I guess the solution now doesn't use any calculus.

A Former Brilliant Member - 4 years, 11 months ago

univariate solution always cool

Son Nguyen - 4 years, 11 months ago

It's nearly to Olympic more than selection test for grade 10.Huhmmmm

Son Nguyen - 4 years, 12 months ago

Will Nesbitt's ineq work here?

Steven Jim - 4 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$