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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
3 months, 1 week 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 · 3 months, 1 week ago

univariate solution always cool · 3 months ago

Is this "mixing of variables"? I think there's some calculus theory behind this method.
Any non calculus approach? · 3 months, 1 week ago

Did you mean the $$f(x, y, z) \geq f(x, y, y)$$? · 3 months ago

Oh no, he changed it, it was $$f(x,y,z) \geq f(x, \sqrt{yz}, \sqrt{yz})$$ I think. · 3 months ago

Ok. I guess the solution now doesn't use any calculus. · 3 months ago

I've to change because later on, I found out that actually, $$f(x,y,z)\leq f(x,\sqrt{yz},\sqrt{yz})$$ · 3 months ago

I've tried but looks like there's no non calculus solution to this · 3 months, 1 week ago

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. · 3 months, 1 week ago

This Inequality was used in this problem,Viet Nam TST 2006 inequality problem: PROBLEM · 3 months ago