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,ca,b,c be positive reals. Find the minimum value of P=a2(a+b)2+b2(b+c)2+c4aP=\dfrac{a^2}{(a+b)^2}+\dfrac{b^2}{(b+c)^2}+\dfrac{c}{4a}.


I've done c4a=c24acc2(c+a)2\dfrac{c}{4a}=\dfrac{c^2}{4ac}\geq\dfrac{c^2}{(c+a)^2}.

I think the minimum value is 34\dfrac34, so the problem is to prove cyca2(a+b)234\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 years, 4 months ago

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1 vote

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Will Nesbitt's ineq work here?

Steven Jim - 2 years, 5 months ago

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Continue the solution of @Ameya Daigavane . Using Cauchy-Schwarz (11+x)2+(11+y)2+(11+z)213(1x+1+1y+1+1z+1)2\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)=1x+1+1y+1+1z+132f(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 xyzx\geq y\geq z, then f(x,y,z)f(x,y,y)f(x,y,z)\geq f(x,y,y).

Now we consider f(x,y,y)f(x,y,y). Since xyx\geq y and xy2=1xy^2=1 we can have y1y\leq 1 and x=1y2x=\frac{1}{y^2}. f(x,y,y)=1x+1+2y+1=y2y2+1+2y+1=y3+3y2+2(y2+1)(y+1)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)32f(x,y,y)\geq\frac{3}{2} y3+3y2+2(y2+1)(y+1)32\frac{y^3+3y^2+2}{(y^2+1)(y+1)}\geq\frac{3}{2} 2y3+6y2+43(y2+1)(y+1)2(y+1)(y2+1)0\Leftrightarrow\frac{2y^3+6y^2+4-3(y^2+1)(y+1)}{2(y+1)(y^2+1)}\geq 0 (1y)32(y+1)(y2+1)0 (True since y1)\Leftrightarrow\frac{(1-y)^3}{2(y+1)(y^2+1)}\geq 0 \ (True \ since \ y\leq 1) So f(x,y,y)32f(x,y,z)32f(x,y,y)\geq\frac{3}{2}\Rightarrow f(x,y,z)\geq\frac{3}{2}, therefore P34P\geq\frac{3}{4}. The equality holds when x=y=z=1x=y=z=1 or a=b=ca=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 - 3 years, 4 months ago

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univariate solution always cool

Laurent Michael - 3 years, 3 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?

Ameya Daigavane - 3 years, 4 months ago

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

Deeparaj Bhat - 3 years, 3 months ago

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

Ameya Daigavane - 3 years, 3 months ago

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

Deeparaj Bhat - 3 years, 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)f(x,yz,yz)f(x,y,z)\leq f(x,\sqrt{yz},\sqrt{yz})

P C - 3 years, 3 months ago

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

P C - 3 years, 4 months ago

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

Laurent Michael - 3 years, 4 months ago

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After setting, x=ba,y=cb,z=acx = \frac{b}{a}, y = \frac{c}{b}, z = \frac{a}{c} this becomes:
For x,y,zR+x, y, z \in R_+ such that xyz=1xyz = 1, (11+x)2+(11+y)2+(11+z)234 \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 years, 4 months ago

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

Laurent Michael - 3 years, 3 months ago

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