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?

## Comments

Sort by:

TopNewestContinue 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

Log in to reply

– Ms Ht · 3 months ago

univariate solution always coolLog in to reply

Any non calculus approach? – Ameya Daigavane · 3 months, 1 week ago

Log in to reply

– Deeparaj Bhat · 3 months ago

Did you mean the \(f(x, y, z) \geq f(x, y, y)\)?Log in to reply

– Ameya Daigavane · 3 months ago

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

– Deeparaj Bhat · 3 months ago

Ok. I guess the solution now doesn't use any calculus.Log in to reply

– Gurīdo Cuong · 3 months ago

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

– Gurīdo Cuong · 3 months, 1 week ago

I've tried but looks like there's no non calculus solution to thisLog in to reply

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

Log in to reply

PROBLEM – Ms Ht · 3 months ago

This Inequality was used in this problem,Viet Nam TST 2006 inequality problem:Log in to reply

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

Log in to reply