11
$\begingroup$

Let $x,y,z,w>0$, and such $x^2+y^2+z^2+w^2=1$. show that $$x+y+z+w+\dfrac{1}{63xyzw}\ge\dfrac{142}{63}\tag{1}$$

I know $$x^2+y^2+z^2+w^2\ge 4\sqrt[4]{x^2y^2z^2w^2}\Longrightarrow xyzw\le \dfrac{1}{16}$$ so we have $$\dfrac{1}{63xyzw}\ge\dfrac{16}{63}$$ But other hand $$(x^2+y^2+z^2+w^2)\ge\dfrac{1}{4}(x+y+z+w)^2\Longrightarrow x+y+z+w\le 2$$ So the above solution is not correct, so how do you prove this inequality $(1)$

$\endgroup$
  • 4
    $\begingroup$ I don't think you can get the estimate by considering $x+y+z+w$ and $\frac{1}{63xyzw}$ separately. Have you tried squaring the whole LHS and then using the AM-GM inequality? $\endgroup$ – MisterRiemann May 17 '18 at 9:17
  • 1
    $\begingroup$ $x+y+z+w+\dfrac{1}{63xyzw}\ge4\sqrt[4]{xyzw}+\dfrac{1}{63xyzw}\ge5\sqrt[5]{\dfrac{1}{63}}$, but this constant is slightly smaller than expected $142/63$. $\endgroup$ – didgogns May 17 '18 at 10:52
  • 2
    $\begingroup$ The expression is minimised when the unknowns are equal. $\endgroup$ – András Salamon May 17 '18 at 19:56
  • $\begingroup$ Who came up with the inequality? Where did you find it? It is very interesting! $(+1) \ \color{orange}{\bigstar}$ $\endgroup$ – Mr Pie May 20 '18 at 6:39
6
+50
$\begingroup$

Let $x=\max\{x,y,z,w\}$ and $f(x,y,z,w)=x+y+z+w+\frac{1}{63xyzw}-\frac{142}{63}.$

Thus, $x\geq\frac{1}{2}$ and $$f(x,y,z,w)-f\left(x,\sqrt{\frac{y^2+z^2+w^2}{3}},\sqrt{\frac{y^2+z^2+w^2}{3}},\sqrt{\frac{y^2+z^2+w^2}{3}}\right)=$$ $$=y+z+w-\sqrt{3(y^2+z^2+w^2)}+\frac{1}{63xyzw}-\frac{1}{63x\left(\sqrt{\frac{y^2+z^2+w^2}{3}}\right)^3}=$$ $$=\frac{\left(\sqrt{\frac{y^2+z^2+w^2}{3}}\right)^3-yzw}{63xyzw\left(\sqrt{\frac{y^2+z^2+w^2}{3}}\right)^3}-\frac{2\sum\limits_{cyc}(y^2-yz)}{\sqrt{3(y^2+z^2+w^2)}+y+z+w)}.$$ We'll prove that $$\frac{\left(\sqrt{\frac{y^2+z^2+w^2}{3}}\right)^3-yzw}{yzw\left(\sqrt{\frac{y^2+z^2+w^2}{3}}\right)^3}\geq\frac{35.7\sum\limits_{cyc}(y^2-yz)}{\left(\sqrt{3(y^2+z^2+w^2)}+y+z+w\right)(y^2+z^2+w^2)^2}.$$ Indeed, let $y+z+w=3p$, $yz+yw+zw=3q^2$ and $yzw=r^3$.

Thus, since $y^2+z^2+w^2=9p^2-6q^2$ does not depend on $r^3$,

the last inequality is equivalent to $g(r^3)\geq0$, where $g$ decreases,

which says that it's enough to prove the last inequality for a maximal value of $r^3$,

which happens for equality case of two variables.

Since the last inequality is homogeneous, it's enough to assume that $z=w=1$

and we need to prove here that $$\frac{\left(\sqrt{(y^2+2)^3}-3\sqrt3y\right)\sqrt{y^2+2}\left(\sqrt{3(y^2+2)}+y+2\right)}{y(y-1)^2}\geq35.7$$ or

$$\frac{((y^2+2)^3-27y^2)\sqrt{y^2+2}\left(\sqrt{3(y^2+2)}+y+2\right)}{y(y-1)^2\left(\sqrt{(y^2+2)^3}+3\sqrt3y\right)}\geq35.7$$ or $$\frac{(y+1)^2(y^2+8)\sqrt{y^2+2}\left(\sqrt{3(y^2+2)}+y+2\right)}{y\left(\sqrt{(y^2+2)^3}+3\sqrt3y\right)}\geq35.7,$$ which is true because $$\min_{y>0}\frac{(y+1)^2(y^2+8)\sqrt{y^2+2}\left(\sqrt{3(y^2+2)}+y+2\right)}{y\left(\sqrt{(y^2+2)^3}+3\sqrt3y\right)}=35.7845...>35.7.$$ Now, $$f(x,y,z,w)-f\left(x,\sqrt{\frac{y^2+z^2+w^2}{3}},\sqrt{\frac{y^2+z^2+w^2}{3}},\sqrt{\frac{y^2+z^2+w^2}{3}}\right)\geq$$ $$\geq\tfrac{35.7\sum\limits_{cyc}(y^2-yz)}{63x\left(\sqrt{3(y^2+z^2+w^2)}+y+z+w\right)(y^2+z^2+w^2)^2}-\tfrac{2\sum\limits_{cyc}(y^2-yz)}{\sqrt{3(y^2+z^2+w^2)}+y+z+w}=$$ $$=\frac{\sum\limits_{cyc}(y^2-yz)(35.7-126x(1-x^2)^2)}{63x\left(\sqrt{3(y^2+z^2+w^2)}+y+z+w\right)(y^2+z^2+w^2)^2}\geq0$$ because $$35.7-126x(1-x^2)^2>0$$ for all $\frac{1}{2}\leq x<1.$

Id est, it's enough to prove the starting inequality for $y=z=w$, which is $$\frac{x+3y}{\sqrt{x^2+3y^2}}+\frac{(x^2+3y^2)^2}{63xy^3}\geq\frac{142}{63}$$ and since the last inequality is homogeneous, we can assume $y=1$ and we need to prove that $$\frac{x+3}{\sqrt{x^2+3}}+\frac{(x^2+3)^2}{63x}\geq\frac{142}{63},$$ which is true.

Done!

It's interesting that $$\min_{x>0}\frac{\frac{x+3}{\sqrt{x^2+3}}+\frac{(x^2+3)^2}{63x}-\frac{142}{63}}{(x-1)^2}=0.00053...,$$ which says that the starting inequality is very strong.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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