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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
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 Gurīdo Cuong · 3 months, 1 week ago

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@Gurīdo Cuong univariate solution always cool Ms Ht · 3 months ago

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

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@Ameya Daigavane Did you mean the \(f(x, y, z) \geq f(x, y, y)\)? Deeparaj Bhat · 3 months ago

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

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@Ameya Daigavane Ok. I guess the solution now doesn't use any calculus. Deeparaj Bhat · 3 months ago

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

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@Ameya Daigavane I've tried but looks like there's no non calculus solution to this Gurīdo Cuong · 3 months, 1 week 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. Ameya Daigavane · 3 months, 1 week ago

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@Ameya Daigavane This Inequality was used in this problem,Viet Nam TST 2006 inequality problem: PROBLEM Ms Ht · 3 months ago

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It's nearly to Olympic more than selection test for grade 10.Huhmmmm Ms Ht · 3 months, 1 week ago

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