Given real numbers $a,b,c >0$ and $a+b+c=3$. Prove that $\frac{a^2}{b}+\frac{b^2}{c} +\frac{c^2}{a} \ge3 + \frac{4}{3}\max\{(a-b)^2;(b-c)^2;(c-a)^2\}$

Question

My first way is to put the inequality under the same degree, so I multipled both sides to $(a+b+c)$, however it leaded to a hard to solve result. Can anyone help me with this problem?

Answer 1

For the current inequality of $ \sum \frac{a^2}{b} \geq 3 + \frac{4}{3} \max ( a-b)^2 $:

1. Replace 3 with $ a+b+c$. The reason for this choice is the "well-known" inequality: $ \sum \frac{ a^2}{b} \geq \sum a $. We're then asking how much more leeway there is in the difference of these terms.

2. We can write the difference as a as Sum of Squares (Figure out how to do this before looking at the hint.). Thus, we want to show that

$$ \sum \frac{ (a-b)^2 } { b} \geq \frac{4}{3} \max ( a-b)^2.$$

This is just Cauchy Schwarz:

• Numerator: $[ \sum | a - b | ] ^2= [2 ( \max (a,b,c) - \min (a, b, c) )]^2 = 4 \max((a-b)^2)$.
• Denominator: $a + b + c = 3$.

Equality holds in the Cauchy Schwarz iff:

• For $ a \geq b \geq c$: $\frac{ a-b}{b} = \frac{b-c}{c} = \frac{a-c}{a} , a+b+c = 3$ $\Rightarrow (a, b, c) = (1, 1, 1) , (\frac{3}{2} , \frac{ 3 \sqrt{5} - 3 } { 4} , \frac{ 9 - 3 \sqrt{5}}{ 4} ). $
• For $ a \geq c \geq b$: $\frac{ a-b}{b} = \frac{c-b}{c} = \frac{a-c}{a} , a+b+c =3 $, which leads to the same cases.

Answer 2

We still can WLOG $a\ge b\ge c$, let $a=c+s+t$ and $b=c+s$, multiply both sides by $3abc$ and write it as

$$(a+b+c)(a^3c+b^3a+c^3b) \ge abc(a+b+c)^2 + 4abc(a-c)^2$$

Left hand side is

$$ct^4+5cst^3+6c^2t^3+s^3t^2+12cs^2t^2+24c^2st^2+13c^3t^2+3s^4t+20cs^3t+45c^2s^2t+43c^3st+15c^4t+2s^5+13cs^4+33c^2s^3+43c^3s^2+30c^4s+9c^5$$

Right hand side is

$$5cst^3+5c^2t^3+17cs^2t^2+28c^2st^2+11c^3t^2+20cs^3t+50c^2s^2t+45c^3st+15c^4t+8cs^4+28c^2s^3+41c^3s^2+30c^4s+9c^5$$

So left minus right is

$$2t^2c^3-2stc^3+2s^2c^3+t^3c^2-4st^2c^2-5s^2tc^2+5s^3c^2+t^4c-5s^2t^2c+5s^4c+s^3t^2+3s^4t+2s^5$$

We know this equals to

$$(c+t+s)(ct-cs-s^2)^2+(c+s)(ct-ts-s^2)^2+c(t^2-cs-s^2)^2\ge 0$$

Or, hiding all the things above, we have

$$(a+b+c)(a^3c+b^3a+c^3b) \ge abc(a+b+c)^2 + 4abc(a-c)^2$$

equals to

$$a(ac-b^2)^2+b(ab-2ca+c^2)^2+c(a^2-2ab+bc)^2\ge 0$$

So, the proof is done.

The equality case is, besides $a=b=c$, we have $c=t=\frac{1+\sqrt 5}2 s$, which means $a=\frac{3}{2}$, $b=\frac{3\sqrt 5-3}{4}$ and $c=\frac{9 -3\sqrt 5}{4}$.

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Below are the proof of $$\frac{a^3}{b}+\frac{b^3}{c} +\frac{c^3}{a} \ge3 + \frac{4}{3}\max\{(a-b)^2;(b-c)^2;(c-a)^2\}$$

And the person who asked this changed the problem.

WLOG $a\ge b\ge c$ (convince yourself that if you exchange $b,c$, the LHS gets larger). Now, brute force.

Let $a=c+s+t$ and $b=c+s$, multiply $abc$ at both sides and making $3$ into $(a+b+c)^2/3$, we have

Left hand side is $a^4c+b^4a+c^4b$, plugging $a=c+s+t$ and $b=c+s$ we have

$$ct^4+4cst^3+4c^2t^3+6cs^2t^2+12c^2st^2+6c^3t^2+s^4t+8cs^3t+18c^2s^2t+16c^3st+5c^4t+s^5+6cs^4+14c^2s^3+16c^3s^2+10c^4s+3c^5.$$

Right hand side is $$abc(\frac{(a+b+c)^2}3+4/3(s+t)^2),$$

which is

$$(5cst^3+5c^2t^3+17cs^2t^2+28c^2st^2+11c^3t^2+20cs^3t+50c^2s^2t+45c^3st+15c^4t+8cs^4+28c^2s^3+41c^3s^2+30c^4s+9c^5)/3$$

You can convince yourself again that LHS minus RHS is all non-negative.

Remark: Playing with Geogebra, $4/3$ can be raised to around $2.445$. The case $4/3$ change to $2$ is also easy.