Prove the inequality knowing $x,y,z \ge 0$ and $xyz=1$ $\frac{x^5}{x^2+1}+\frac{y^5}{y^2+1}+\frac{z^5}{z^2+1}\ge\frac{3}{2}$

Question

Prove the inequality knowing $x,y,z \ge 0$ and $xyz=1$ $$\frac{x^5}{x^2+1}+\frac{y^5}{y^2+1}+\frac{z^5}{z^2+1}\ge\frac{3}{2}$$

Starting from the condition set that $xyz =1$ I did this : $$x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}$$ $$x^3+y^3+z^3\ge3$$ I also know that the fraction $\frac{3}{2}$ usually either comes from the famous $Nesbitt$ equation or from some sort of $Tittu$ that leads to $\frac{9}{4}$ which then is simplified.

However I can't find any way how to manipulate the inequality. I did try to multiply and rewrite the inequality so that $x,y,z$ are an even power like this : $$\frac{x^4}{x^2+yz}+\frac{y^4}{y^2+xz}+\frac{z^4}{z^2+xy}\ge\frac{3}{2}$$

However this didn't bring me anywhere. Any kind of help is welcomed.

Answer 1

We use the tangent line method, ie first set $f(x)= \frac{x^5}{x^2+1}$ so that we have to show that $$f(x)+f(y)+f(z) \geq \frac{3}{2}$$

Note $f’’(x) \geq 0 \ \forall x \in \mathbb R_{\geq 0}$. Considering the tangent to $f(x)$ at $x=1$, it becomes clear that $f(x) \geq 2x-\frac{3}{2}$ (try to verify this by cross multiplication) so that $$f(x)+f(y)+f(z) \geq 2(x+y+z) - \frac{9}{2} \geq 6-\frac{9}{2} = \frac32$$ where the last inequality follows via AM-GM. Equality is attained at $(1,1,1)$.

Answer 2

Since we have the constraint $xyz = 1$, we can make the substitution $x = \frac{a}{b}$, $y = \frac{b}{c}$, and $z = \frac{c}{a}$, where $a$, $b$, and $c$ are positive real numbers. (If you haven't seen this before, you can for example let $a = x$, $b = 1$, and $c = \frac{1}{y}$, but we aren't really interested in the particular values of $a$, $b$, and $c$.)

We then want to prove that the inequality $$ \frac{a^5}{a^2 b^3 + b^5} + \frac{b^5}{b^2 c^3 + c^5} + \frac{c^5}{c^2 a^3 + a^5} \geq \frac{3}{2} $$ holds for all positive real numbers $a$, $b$, and $c$ (without any additional constraints) or, equivalently, $$ \frac{a^6}{a^3 b^3 + a b^5} + \frac{b^6}{b^3 c^3 + b c^5} + \frac{c^6}{c^3 a^3 + c a^5} \geq \frac{3}{2}. $$

By the Cauchy-Schwarz Inequality (sometimes called Titu's Lemma, like in your question) $$ \frac{a^6}{a^3 b^3 + a b^5} + \frac{b^6}{b^3 c^3 + b c^5} + \frac{c^6}{c^3 a^3 + c a^5} \geq \frac{(a^3 + b^3 + c^3)^2}{a^3 b^3 + b^3 c^3 + c^3 a^3 + ab^5 + bc^5 + ca^5}, $$ so it is enough to show that $$ 2a^6 + 2b^6 + 2c^6 + 4a^3 b^3 + 4b^3 c^3 + 4c^3 a^3 \geq 3a^3 b^3 + 3b^3 c^3 + 3c^3 a^3 + 3ab^5 + 3bc^5 + 3ca^5, $$ or more simply $$ 2a^6 + 2b^6 + 2c^6 + a^3 b^3 + b^3 c^3 + c^3 a^3 \geq 3ab^5 + 3bc^5 + 3ca^5. $$

Now by AM-GM, we have $$ a^6 + a^6 + c^3 a^3 \geq 3ca^5 $$ and similarly $$ 2b^6 + a^3 b^3 \geq 3ab^3 \quad\text{ and }\quad 2c^6 + b^3 c^3 \geq 3bc^5, $$ and adding these together gives us the desired inequality.

Answer 3

Alternate approach:
We try to homogenize the inequality. Let $(xyz)^{2/3}$ be denoted by $p$. $$\begin{align} \sum_{cyc} \frac{x^5}{x^2+1} &= \sum_{cyc}\frac{x^5}{x^2+p} \\ &= \frac1{xyz}\sum_{cyc}\frac{x^5}{x^2+p} \\ &\ge \frac 3{x^3+y^3+z^3} \sum_{cyc}\frac{x^6}{x^3+xp} \tag{AM-GM} \\ &\ge \frac 3{\sum_{cyc} x^3} \frac{\left(\sum_{cyc} x^3\right)^2}{\sum_{cyc} x^3+xp} \tag{C-S} \\ &= 3\cdot \frac{\sum_{cyc} x^3}{\sum_{cyc} x^3+xp} \ge \frac{3}{2} \end{align}$$ where the last inequality follows since $$\sum_{cyc}x^3 \ge \sum_{cyc} xp$$which is trivially true by Muirhead or AM-GM.

Answer 4

$$\frac{x^5}{x^2 +1} = \frac{4x^3-1}{6} +\underbrace{ \frac{(x-1)^2(2x^3+4x^2+2x+1)}{6(x^2+1)}}_{\ge0} \ge \frac{4x^3-1}{6}$$ Then $$\sum_{\text{sym}}\frac{x^5}{x^2 +1} \ge \sum_{\text{sym}}\frac{4x^3-1}{6} = -\frac{1}{2} + \underbrace{\frac{2}{3}\sum_{\text{sym}}x^3}_{\ge 2xyz = 2} \ge \frac{3}{2}$$

The equality occurs if and only if $x = y= z= 1$

Answer 5

Another solution: $$\sum \frac{x^5}{x^2+1} =\frac{1}{(xyz)^{2/3}}\cdot \sum \frac{x^5}{x^2+xyz} =$$

$$= \frac{1}{(xyz)^{2/3}}\cdot \sum \frac{x^4}{x+yz} \overset{\text{AM-GM}}\ge \frac{3}{x^2+y^2+z^2}\cdot\sum \frac{(x^2)^2}{x+yz} \ge $$

$$\overset{\text{Titu}}\ge \frac{3}{x^2+y^2+z^2}\cdot \frac{(x^2+y^2+z^2)^2}{xy+yz+zx+x+y+z}\ge $$

$$\overset{*}\ge \frac{3(x^2+y^2+z^2)}{x^2+y^2+z^2+x^2+y^2+z^2} = \frac32.$$

$(*)$ is true because $$x^2+y^2+z^2\ge xy+yz+zx$$ and $$x^2+y^2+z^2\ge x+y+z$$ which follows from

$$x^2+y^2+z^2\ge \frac{(x+y+z)^2}{3} \ge (x+y+z)\sqrt[3]{xyz}=x+y+z.$$

Answer 6

No tricks requires, just standard techniques.

Lemma: Apply Jensen's to show that

If $ \frac{ x+y+z}{3} = A $, then $\sum \frac{ x^5 } { x^2 + 1 } \geq 3 \frac{ A^5}{A^2 + 1 }$.

Proof: Work through the tedious verification that $f''(x) = \frac{ 2x^3 (3x^4 + 9x^2 + 10 } { (x^2+1)^3 } \geq 0$.

Corollary: The condition $ xyz=1 $ tells us that $ x+y+z \geq 3$. Hence, $$ \sum \frac{ x^5 }{ x^2 + 1 } \geq 3 \frac{ A^ 5 } { A^2 + 1 } \geq \frac{3}{2}.$$

Equality holds iff $ A = 1, x=y=z = 1$.

Answer 7

From the hypothesis, we can rewrite the first inequality as: $$\frac{x^4}{x+yz}+\frac{y^4}{y+zx}+\frac{z^4}{z+xy} \ge \frac{3}{2}$$ By applying Titu's lemma, we have: $$\frac{x^4}{x+yz}+\frac{y^4}{y+zx}+\frac{z^4}{z+xy} \ge \frac{(x^2+y^2+z^2)^2}{x+y+z+xy+yz+xz} \ge \frac{\dfrac{1}{9}(x+y+z)^4}{x+y+z+\frac{1}{3}(x+y+z)^2}$$ Let $x+y+z=t$.Thus, we just need to prove: $$\frac{t^4}{9t+3t^2} \ge \frac{3}{2}$$ Which is equivalent to: $$t(t-3)(2t^2+6t+9) \ge 0$$ It's obvious since $t=a+b+c \ge 3(abc)^{\frac{1}{3}}=3$