Prove that: $\frac{1}{1+x+x^2+x^3}+\frac{1}{1+y+y^2+y^3}+\frac{1}{1+z+z^2+z^3}+\frac{1}{1+t+t^2+t^3}\ge1$

Question

Let $xyzt$ st. $xyzt=1 $ Prove that: $$\dfrac{1}{1+x+x^2+x^3}+\dfrac{1}{1+y+y^2+y^3}+\dfrac{1}{1+z+z^2+z^3}+\dfrac{1}{1+t+t^2+t^3}\ge1$$ It's look like Vasc inequality (Let $a,b,c>0$ st. $abc=1$ prove $\dfrac{1}{1+a+a^2}+\dfrac{1}{1+b+b^2}+\dfrac{1}{1+c+c^2}\ge1$) so I did the same to prove this Vasc inequality.

$\bullet$ Let $(x;y;z;t)=(\dfrac{ab}{c^2};\dfrac{bc}{d^2};\dfrac{cd}{a^2};\dfrac{da}{b^2})$

$\bullet$ We need to prove $$ \dfrac{a^6}{a^6+bcd^6+b^2c^2d^2+b^3c^3}+\dfrac{b^6}{b^6+dab^4+a^2a^2b^2+d^3a^3}+\dfrac{c^6}{c^6+abc^4+a^2b^2c^2+a^3b^3}+\dfrac{d^6}{d^6+bcd^4+b^2c^2d^2+b^3c^3} \ge 1$$

$\bullet$ By Cauchy-Schwarz, we have : $$\dfrac{a^6}{a^6+bcd^6+b^2c^2d^2+b^3c^3}+\dfrac{b^6}{b^6+dab^4+a^2a^2b^2+d^3a^3}+\dfrac{c^6}{c^6+abc^4+a^2b^2c^2+a^3b^3}+\dfrac{d^6}{d^6+bcd^4+b^2c^2d^2+b^3c^3} \ge\dfrac{(a^3+b^3+c^3+d^3)^2}{ a^6+bcd^6+b^2c^2d^2+b^3c^3+b^6+dab^4+a^2a^2b^2+d^3a^3+c^6+abc^4+a^2b^2c^2+a^3b^3+d^6+bcd^4+b^2c^2d^2+b^3c^3}$$

$\bullet$ The problem is:$(a^3+b^3+c^3+d^3)^2\ge a^6+bcd^6+b^2c^2d^2+b^3c^3+b^6+dab^4+a^2a^2b^2+d^3a^3+c^6+abc^4+a^2b^2c^2+a^3b^3+d^6+bcd^4+b^2c^2d^2+b^3c^3$

or

$$a^3b^3+a^3c^3+a^3d^3+b^3c^3+b^3d^3+c^3d^3 \ge abc^4+a^2b^2c^2+bcd^4+b^2c^2d^2+cda^4+c^2d^2a^2+dab^4+d^2a^2b^2$$

but I can't prove it, maybe it isn't true ? Pls help me

Answer 1

It's wrong. Try $x\rightarrow-1^-$.

For positive variables let $x=e^a$, $y=e^b$, $z=e^c$,$z=e^d$ and $f(x)=\frac{1}{(e^x+1)(e^{2x}+1)}.$

Thus, $$a+b+c+d=0$$ and $$f''(x)=\frac{e^x(9e^{5x}+11e^{4x}+6e^{3x}-6e^{2x}-3e^x-1)}{(e^x+1)^3(e^{2x}+1)^3}>0$$ for any $x\geq0=\frac{a+b+c+d}{4}.$

Thus, by the Vasc's RCF Theorem it's enough to prove our inequality for equality case of three variables.

Can you end it now?

About RCF see here: https://kheavan.files.wordpress.com/2010/06/2007_1_applic.pdf

I got that finally we need to prove that: $$(x-1)^2(2x^4+x^3+x+2)\geq0,$$ which is obvious.

Answer 2

Hint for a different approach (due to Vasile Cirtoaje, Algebraic Inequalities - Old and New Method, p. 393).

Show first that for any positive $u,v$, $$\frac{1}{1+u^2+u^4+u^6}+\frac{1}{1+v^2+v^4+v^6}\geq \frac{1}{1+u^3v^3}.$$ Then $$\frac{1}{1+x+x^2+x^3}+\frac{1}{1+y+y^2+y^3}\geq \frac{1}{1+\sqrt{x^3y^3}}$$ and $$\frac{1}{1+z+z^2+z^3}+\frac{1}{1+t+t^2+t^3}\geq \frac{1}{1+\sqrt{z^3t^3}}= \frac{\sqrt{x^3y^3}}{\sqrt{x^3y^3}+1}.$$ and after summing them, your inequality easily follows.