$a,b,c\in\mathbb R^+$, prove $\frac{a^3}{bc} +\frac{b^3}{ca} + \frac{c^3}{ab} \geq \frac{a^2 + b^2}{2c} + \frac{b^2 + c^2}{2a} + \frac{c^2 + a^2}{2b}$

Question

$a$,$b$,$c \in \mathbb R^+$, prove that $$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq \frac{a^2 + b^2}{2c} + \frac{b^2 + c^2}{2a} + \frac{c^2 + a^2}{2b}$$

My Attempt :

$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq \frac{a^2 + b^2}{2c} + \frac{b^2 + c^2}{2a} + \frac{c^2 + a^2}{2b}$

or, $a^4 + b^4 + c^4 \ge \frac{ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)}{2}$

or, $a^4 + b^4 + c^4 \ge \frac{(a^2+b^2+c^2)(ab+bc+ca)}{2} \ge \frac{ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)}{2}$

or, $a^4 + b^4 + c^4 \ge \frac{(a^2+b^2+c^2)(ab+bc+ca)}{2}$

How to proceed further?

And also tell if there's an more elegant way to it using rearrangement or Chebyshev!

Answer 1

Hint:

Check if $${a^3\over b} + {b^3\over a}\geq a^2+b^2$$ is true,

i.e. $$a^4+b^4\geq ab(a^2+b^2)$$

i.e. $$(a-b)(a^3-b^3)\geq 0 $$

Answer 2

Without loss of generality we can assume that $0 < a \le b \le c$. Then $$ a^4+ b^4 + c^4 \ge a^3 b + b^3 c + c^3 a \\ a^4+ b^4 + c^4 \ge a b^3 + b c^3 + c a^3 \\ $$ using the rearrangement inequality twice. Adding these two inequalities and dividing by $2abc$ gives the desired result.

Answer 3

By rearranging the inequality, you have :

$$ \begin{align}\sum_{\rm{cyc}} \left(\frac{a^3}{bc} + \frac{b^3}{ca}\right)&≥\sum_{\rm{cyc}}\left( \frac {a^2+b^2}{c}\right)\end{align} $$

Then, multiplying both side of the inequality by $abc≠0$, we have :

$$ \begin{align}a^4+b^4&≥\frac {\left(a^2+b^2\right)^2}{2}\\ &≥ab\left(a^2+b^2\right) \tag {*}\end{align} $$

Done .

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$(^{*})\thinspace\thinspace\thinspace\rm{Comment:}$

The following inequality

$$a^4+b^4≥\frac {\left(a^2+b^2\right)^2}{2}$$

or equivalently,

$$(a^4+b^4)(1+1)≥\left(a^2+b^2\right)^2$$

can be simply derived from the Cauchy–Schwarz inequality .