This problem posted again at http://www.mathlinks.ro/Forum/viewtopic.php?t=2488 I rewrite the solution in $\text{\LaTeX}$ Lagrangia wrote: this is siuhohung's solution 4 the 1st problem: Divide both sides by $u^{4}$ then the inequality becomes $(1-4\frac{v}{u}\frac{w}{u})^{2}> K(2(\frac{v}{u})^{2}-\frac{w}{u})(2(\frac{w}{u})^{2}-\frac{w}{u}))$ Sub. $a=\frac{v}{u}, b=\frac{w}{u}$ $(1-4ab)^{2}>K(2a^{2}-b)(2b^{2}-a)$ The constaint given is transformed to $ab<\frac{1}{4}$ Then when $a =\frac{1}{2}\Rightarrow b<\frac{1}{2}$ and when $b=\frac{1}{2}\Rightarrow a<\frac{1}{2}$ Therefore $(a-b)^{2}+(a-\frac{1}{2})^{2}+(b-\frac{1}{2})^{2}>0$ Expand and simplify, we have $(2a^{2}-b)+(2b^{2}-a)>2ab-\frac{1}{2}$ Let $x=2a^{2}-b$ $y=2b^{2}-a$ $z=1-4ab$ So we are having, $x+y>-\frac{z}{2}$ and $z^{2}>Kxy$ $2(x+y)>-z$ $4(x+y)^{2}>z^{2}>Kxy$ $4x^{2}+(8-K)xy+4y^{2}>0$ Obviously, $K=16$

WLOG $uw-2v^2>0,uv-2w^2>0$, hence $16(2v^{2}-uw)(2w^{2}-uv)\le4 \left((uw-2v^2)+(uv-2w^2)\right)^2$ $=\left(2u(v+w)-4(v^2+w^2)\right)^2\le\left(u^2+(v+w)^2-4(v^2+w^2)\right)^2$ $=\left(u^2+2vw-3(v^2+w^2)\right)^2\le\left(u^2+2vw-6vw\right)^2=(u^{2}-4vw)^{2}.$

Lagrangia wrote: $2(x+y)>-z$ $4(x+y)^{2}>z^{2}$ Why?

sqing wrote: WLOG $uw-2v^2>0,uv-2w^2>0$ What happens if $uw-2v^2<0$ and $uv-2w^2<0$?