# 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? Note by Tran Quoc Dat
2 years, 11 months ago

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## Comments

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Will Nesbitt's ineq work here?

- 2 years ago

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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

- 2 years, 11 months ago

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univariate solution always cool

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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Oh no, he changed it, it was $$f(x,y,z) \geq f(x, \sqrt{yz}, \sqrt{yz})$$ I think.

- 2 years, 11 months ago

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Ok. I guess the solution now doesn't use any calculus.

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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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.

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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