2
$\begingroup$

let $x,y,z,w\in R$,and such $x+y+z+w=2$.show that $$\sum_{cyc}\dfrac{x}{x^2-x+1}\le\dfrac{8}{3}$$

I have only solve when $x,y,z,w>0$, because $$\dfrac{x}{x^2-x+1}\le\dfrac{4}{3}x$$ so $$\sum_{cyc}\dfrac{x}{x^2-x+1}\le\sum\dfrac{4x}{3}=\dfrac{8}{3}$$

$\endgroup$

marked as duplicate by Yanior Weg, Jyrki Lahtonen, Cesareo, Zestylemonzi, eyeballfrog May 11 at 21:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ So... what is the question? $\endgroup$ – Simply Beautiful Art May 10 at 2:05
  • $\begingroup$ but for $x,y,z,w$ be real numbers,my method can't works $\endgroup$ – function sug May 10 at 2:17
  • $\begingroup$ Perhaps you should emphasize that at the end to make the question clearer. $\endgroup$ – Simply Beautiful Art May 10 at 2:19
1
$\begingroup$

Let $x=\frac{a}{2},$ $y=\frac{b}{2},$ $z=\frac{c}{2}$ and $w=\frac{d}{2}.$

Thus, $a+b+c+d=4$ and we need to prove that $$\sum_{cyc}\frac{a}{a^2-2a+4}\leq\frac{4}{3}$$ or $$\sum_{cyc}\frac{2a}{a^2-2a+4}\leq\frac{8}{3}$$ or $$ \sum_{cyc}\left(\frac{2a}{a^2-2a+4}-1\right)\leq\frac{8}{3}-4$$ or $$\sum_{cyc}\frac{(a-2)^2}{a^2-2a+4}\geq\frac{4}{3}.$$ Now, by C-S $$\sum_{cyc}\frac{(a-2)^2}{a^2-2a+4}=\sum_{cyc}\frac{(a-2)^4}{(a^2-2a+4)(a-2)^2}\geq\frac{\left(\sum\limits_{cyc}(a-2)^2\right)^2}{\sum\limits_{cyc}(a^2-2a+4)(a-2)^2}.$$ Hence, it's enough to prove that $$3\left(\sum\limits_{cyc}(a-2)^2\right)^2\geq4\sum\limits_{cyc}(a^2-2a+4)(a-2)^2$$ or $$3\left(\sum_{cyc}a^2\right)^2\geq4\sum_{cyc}(a^4-6a^3+16a^2-8).$$ Now, let $a+b+c+d=4u$, $ab+ac+ad+bc+bd+cd=6v^2,$ where $v^2$ can be negative,

$abc+abd+acd+bcd=4w^3$ and $abcd=t^4,$ where $t^4$ can be negative.

Thus, we need to prove that $$3(16u^2-12v^2)^2\geq4(256u^4-384u^2v^2+72v^4+64uw^3-4t^4)-24(64u^3-72uv^2+12w^3)u+64(16u^2-12v^2)u^2-128u^4$$ or $$24u^4-36u^2v^2+9v^4+2uw^3+t^4\geq0.$$ Now, $$(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2\geq0$$ gives $$t^4\geq4uw^3-3v^4.$$ Thus, it's enough to prove that $$4u^4-6u^2v^2+v^4+uw^3\geq0.$$ Now, $a$, $b$, $c$ and $d$ are real roots of the equation $$(x-a)(x-b)(x-c)(x-d)=0$$ or $$x^4-4ux^3+6u^2x^2-4w^3x+t^4=0,$$ which by the Rolle's theorem gives that the equation $$ (x^4-4ux^3+6u^2x^2-4w^3x+t^4)'=0$$ or $$x^3-3ux^2+3v^2x-w^3=0$$ has three real roots, which says that there are reals $p$, $q$ and $r$, for which $$3u=p+q+r,$$ $$3v^2=pq+pr+qr$$ and $$pqr=w^3.$$ Id est, it's enough to prove that $$4u^4-6u^2v^2+v^4+uw^3\geq0$$ for three variables.

But this inequality is a linear inequality of $w^3,$ which says that it's enough to prove it for the extreme value of $w^3$, which happens for equality case of two variables.

Let $p=q$.

If $p=q=0$, the inequality is obviously true.

But for $p=q\neq0$ we can assume that $p=q=1$ and it's enough to prove that $$\frac{4(r+2)^4}{81}-\frac{2(r+2)^2(2r+1)}{9}+\frac{(2r+1)^2}{9}+\frac{(r+2)r}{3}\geq0$$ or $$(r-1)^2(2r+1)^2\geq0$$ and we are done!

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.