# prove this inequality with $63$

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 $(1)$

• 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? – MisterRiemann May 17 '18 at 9:17
• $x+y+z+w+\dfrac{1}{63xyzw}\ge4\sqrt[4]{xyzw}+\dfrac{1}{63xyzw}\ge5\sqrt[5]{\dfrac{1}{63}}$, but this constant is slightly smaller than expected $142/63$. – didgogns May 17 '18 at 10:52
• The expression is minimised when the unknowns are equal. – András Salamon May 17 '18 at 19:56
• Who came up with the inequality? Where did you find it? It is very interesting! $(+1) \ \color{orange}{\bigstar}$ – Mr Pie May 20 '18 at 6:39

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.