Inequality $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5} $

Question

I have trouble with solving this inequality: Prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5}$ for a,b,c>0.

Using Cauchy-Schwartz I got this: $\frac{a^3}{3b^3+2c^3} +\frac{b^3}{3c^3+2a^3} +\frac{c^3}{3a^3+2b^3} \geq \frac{3}{5} $

Now I have to prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{a^3}{3b^3+2c^3} +\frac{b^3}{3c^3+2a^3} +\frac{c^3}{3a^3+2b^3}$

Any help is appreciated!

Answer 1

By the Cauchy-Schwartz inequality, we have:

$$(\frac{a^3}{3ab^2+2c^3}+\frac{b^3}{3bc^2+2a^3}+\frac{c^3}{3ca^2+2b^3})(a(3ab^2+2c^3)+b(3bc^2+2a^3)+c(3ca^2+2b^3)) \\ \geq (a^2+b^2+c^2)^2 \\ \implies \frac{a^3}{3ab^2+2c^3}+\frac{b^3}{3bc^2+2a^3}+\frac{c^3}{3ca^2+2b^3} \geq \frac{(a^2+b^2+c^2)^2}{3(a^2b^2+a^2c^2+b^2c^2)+2(ac^3+ba^3+cb^3)}.$$ Therefore, we are done if we show that:

$$\frac{(a^2+b^2+c^2)^2}{3(a^2b^2+a^2c^2+b^2c^2)+2(ac^3+ba^3+cb^3)} \geq \frac{3}{5},$$

or equivalently:

$$5(a^4+b^4+c^4)+(a^2b^2+a^2c^2+b^2c^2)\geq 6(ac^3+ba^3+cb^3).$$

However, we already have: $$a^4+a^2b^2 \geq 2ba^3 \\b^4+b^2c^2 \geq 2cb^3 \\c^4+a^2c^2 \geq 2ac^3 .$$

Thus, it is enough to show that:

$$a^4+b^4+c^4 \geq ac^3+ba^3+cb^3,$$

which is a consequence of the rearrangement inequality (note that $a,b,c$ and $a^3,b^3,c^3$ are sorted in the same way).

Answer 2

By Am-Gm inequality we have $$3ab^2\leq a^3+b^3+b^3$$ so $$3ab^2 +2c^3\leq a^3+2b^3+2c^3$$ so $${a^3 \over 3ab^2+2c^3} \geq {a^3\over a^3+2b^3+2c^3}$$ Let $x=a^3$, $y=b^3$ and $z=c^3$. We can assume that $x+y+z=1$, and so we have to prove $$f(x) +f(y) +f(z)\geq 3/5$$ where $$f(x) ={x\over 2-x}$$ The tangent at $x=1/3$ is $$t(x)={18\over 25}x- {1\over 25}$$ Since $f(x) \geq t(x)$ for all $x\in[0,1)$ we have $$f(x) +f(y) +f(z)\geq t(x) +t(y) +t(z) = 3/5$$

Answer 3

Because by C-S $$\sum_{cyc}\frac{a^3}{3ab^2+2c^3}-\frac{3}{5}=\sum_{cyc}\frac{a^4}{3a^2b^2+2c^3a}-\frac{3}{5}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(3a^2b^2+2c^3a)}-\frac{3}{5}=$$ $$=\frac{\sum\limits_{cyc}(5a^4-6a^3b+a^2b^2)}{5\sum\limits_{cyc}(3a^2b^2+2c^3a)}=\frac{\sum\limits_{cyc}\left(5a^4-6a^3b+a^2b^2-(a^4-b^4)\right)}{5\sum\limits_{cyc}(3a^2b^2+2c^3a)}=$$ $$=\frac{\sum\limits_{cyc}(a-b)^2(4a^2+2ab+b^2)}{5\sum\limits_{cyc}(3a^2b^2+2c^3a)}\geq0.$$

Answer 4

Now I have to prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{a^3}{3b^3+2c^3} +\frac{b^3}{3c^3+2a^3} +\frac{c^3}{3a^3+2b^3}$

Take $a=10$, $b=c=1$. Is it true?

From nonusers answer, assuming $a^3+b^3+c^3=1$ due to homogeneity,

$$\sum_{cyc}{a^3 \over 3ab^2+2c^3} \geq \sum_{cyc}{a^3\over a^3+2b^3+2c^3}=\sum_{cyc}{a^3 \over 2-a^3}\geq \frac35$$ since minumum happens when $a^3=b^3=c^3=\frac13$ by an easy application of method of Lagrange's multipliers.