Prove $\frac{x^3+2}{2+x+y+z^3}+\frac{y^3+2}{2+y+z+x^3}+\frac{z^3+2}{2+z+x+y^3}\geq\frac{9}{5}$

Question

Prove $$\frac{x^3+2}{2+x+y+z^3}+\frac{y^3+2}{2+y+z+x^3}+\frac{z^3+2}{2+z+x+y^3}\geq\frac{9}{5}$$ for $x,y,z>0$.

My work:

Using AM-GM: $\sqrt{z^3} \leq \frac{z^3+2}{3}$

$z^3+2\geq3z$

Now I have $$\frac{x}{2+x+y+z^3}+\frac{y}{2+y+z+x^3}+\frac{z}{2+z+x+y^3}\geq\frac{3}{5}$$ and I don't know what to do next.

Any help is appreciated!

Answer 1

It is clear that equality is attained at least at $x=y=z=1$. Therefore, any subsequent inequalities have to respect this. Furthermore, note that we should not change the degrees of the terms as present in such a manner that as $x,y,z \to 0^+$ or $x,y,z \to +\infty$, the simplified inequality fails.

The linked post tells us how to proceed carefully. Indeed, if we use the AM-GM inequality with $x^3,1,1$, we get $\frac{x^3+2}{3} \geq x$ for any $x>0$. Recognize that equality is attained when $x^3=1$ i.e. when $x=1$. Adding any arbitrary positive real $r>0$ and inverting that inequality, we get $\frac{1}{\frac{x^3+2}{3}+r} \leq \frac{1}{x+r}$.

Perhaps it is important to highlight why this is being attempted. Indeed, as it stands, there are terms in the denominator, namely $x,y$ whose degrees do not match up to that of the numerator and the rest of the terms. If we can obtain an inequality which does this, then it may be possible to tackle the resulting inequality through typical methods for homogenous inequalities.

For example, take $\frac{x^3+2}{2+x+y+z^3}$. Applying the AM-GM inequalities for both $x^3,1,1$ and $y^3,1,1$ and using the trick I mentioned above yields $$ \frac{x^3+2}{2+x+y+z^3} \geq \frac{x^3+2}{(2+z^3)+\frac{x^3+2}{3}+\frac{y^3+2}{3}} $$ Write $a=2+x^3,b=2+y^3,c=2+z^3$. Then, what we have is that $$ \frac{x^3+2}{2+x+y+z^3} \geq \frac{a}{c+\frac a3+\frac b3} = \frac {3a}{3c+a+b} $$ Applying this inequality for the other terms as well, $$ \frac{x^3+2}{2+x+y+z^3} +\frac{y^3+2}{2+y+z+x^3}+\frac{z^3+2}{2+z+x+y^3} \\ \geq \frac{3a}{3c+a+b}+\frac{3b}{3a+b+c}+\frac{3c}{3b+c+a} $$

This is familiar territory for those who have seen Nesbitt's inequality and variants, and there are a couple of ways that one can finish from here.

My favorite one would be the following : factorize the $3$ out, then multiply the top and bottom of the first term by $a$, the second by $b$ and the third by $c$ to get $$ \frac{3a}{3c+a+b}+\frac{3b}{3a+b+c}+\frac{3c}{3b+c+a} \\ = 3\left(\frac{a^2}{3ac+a^2+ab}+\frac{b^2}{3ab+b^2+bc}+\frac{c^2}{3bc+c^2+ac} \right) $$

We will now use Titu's lemma, a popular consequence of Cauchy-Schwarz. It states that if $x_1,\ldots,x_n$ are real numbers and $a_1,\ldots,a_n$ are positive reals, then $\sum \frac{x_i^2}{a_i} \geq \frac{(\sum x_i)^2}{\sum a_i}$, where equality holds if and only if $x_i = ka_i$ for some non-zero real constant $k$ independent of $i$. It is easy to observe that when $x=y=z=1$, $a=b=c=3$ and equality is attained in this inequality.

Applying Titu's lemma with the obvious $x_i,a_i$,$$ 3\left(\frac{a^2}{3ac+a^2+ab}+\frac{b^2}{3ab+b^2+bc}+\frac{c^2}{3bc+c^2+ac} \right) \\ \geq 3\left(\frac{(a+b+c)^2}{a^2+b^2+c^2+4(ab+bc+ca)}\right) \tag{1} \label{1} $$

We will be done if we can somehow compare the bottom to the top of the RHS, which amounts to comparing the denominator when $4$ is changed by $2$. Note that when $a=b=c$, we expect to see equality in such an inequality.

Note that we are comparing $a^2+b^2+c^2+4(ab+bc+ca)$ to $a^2+b^2+c^2+2(ab+bc+ca)$. Therefore, it's a comparison of homogenous quantities, but basically we have to "shift" the $ab+bc+ca$ coefficient onto the $a^2+b^2+c^2$ coefficient. The only way we know of doing that is using the tight inequality $$ ab+bc+ca \leq a^2+b^2+c^2 $$ Multiply this by $\epsilon$ where $\epsilon$ will be decided later, and use it to get $$ a^2+b^2+c^2+4(ab+bc+ca) \leq (1+\epsilon)(a^2+b^2+c^2)+(4-\epsilon)(ab+bc+ca) $$ we want to make the right hand side into a constant multiple of $(a+b+c)^2$. To do this, we need $2(1+\epsilon) = (4-\epsilon)$. This leads to $\epsilon = \frac{2}{3}$ whence $$ (1+\epsilon)(a^2+b^2+c^2)+(4-\epsilon)(ab+bc+ca) = (1+\epsilon)(a+b+c)^2 \\ \implies a^2+b^2+c^2+4(ab+bc+ca) \leq \frac{5}{3}(a+b+c)^2 $$

From \eqref{1}, $$ 3\left(\frac{(a+b+c)^2}{a^2+b^2+c^2+4(ab+bc+ca)}\right) \geq 3\left(\frac{(a+b+c)^2}{\frac 53 (a+b+c)^2}\right) \geq \frac 95 $$

which, along with the inequalities preceding \eqref{1}, finish the proof.