# Inequality : $\sqrt{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt{3}$

Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt{3}.$$

From Micheal Rozenberg's answer :

$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq\sqrtx\left(\sqrt{(x^3+2)^2}+\sqrt{3x(x^3+2)}+\sqrt{9x^2}\right)$,

Prove that $(x+2)^2(x+2\sqrt{x}+3)\geq9\sqrt{x(x^3+2)^2}$,

LHS :

$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq (x+2) \frac{x+2} {\sqrt{3}}\left(\frac{x+2}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)= \frac{(x+2)^2}{3}(x+2\sqrt{x}+3)$

RHS :

$\sqrt{3x(x^3+2)}\leq \sqrt{(x^3+2)^2}$

$\sqrt{9x^2}\leq \sqrt{(x^3+2)^2}$

so $\sqrt{(x^3+2)^2}+\sqrt{3x(x^3+2)}+\sqrt{9x^2}\leq 3\sqrt{(x^3+2)^2}$

$\sqrt{x}(\sqrt{(x^3+2)^2}+\sqrt{3x(x^3+2)}+\sqrt{9x^2})\leq 3\sqrt{x(x^3+2)^2}$

Thus,

$\frac{(x+2)^2}{3}(x+2\sqrt{x}+3) \geq 3\sqrt{x(x^3+2)^2}$

$(x+2)^2(x+2\sqrt{x}+3) \geq 9\sqrt{x(x^3+2)^2}$

• yuck.................. – K Split X Aug 9 '17 at 3:27

## 4 Answers

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $$\sqrt{\frac{27u^2-27uv^2}{w^3}+3}+\sqrt{\frac{v^2}{3u^2-2v^2}}\geq1+\sqrt3,$$ which is $f(w^3)\geq0,$ where $f$ is a decreasing function.

Thus, it's enough to prove our inequality for a maximal value of $w^3$.

Now, $x$, $y$ and $z$ are positive roots of the following equation. $$(X-x)(X-y)(X-z)=0$$ or $$X^3-3uX^2+3v^2X-w^3=0$$ or $$X^3-3uX^2+3v^2X=w^3,$$ which says that the graph of $f(X)= X^3-3uX^2+3v^2X$ and the line $Y=w^3$

have three common points $(x,f(x))$, $(y,f(y))$ and $(z,f(z)).$

Now, draw the graph of $f$.

Indeed, $f(X)= X^3-3uX^2+3v^2X$, which gives $$f'(X)=3X^2-6uX+3v^2=3(X^2-2uX+v^2)=$$ $$=3\left(X-(u+\sqrt{u^2-v^2})\right)\left(X-(u-\sqrt{u^2-v^2})\right).$$

Thus, $X_{max}=u-\sqrt{u^2-v^2}$, $X_{min}=u+\sqrt{u^2-v^2}$

and the graph of $f$ goes through origin $(0,0)$.

Draw it, please!

Let $z\leq y\leq x$, $u$ and $v^2$ be constants and $w^3$ increases.

Hence, $x$, $y$ and $z$ changes and $w^3$ will get a maximal value,

when a line $Y=w^3$ will touch to the graph of $f$ in the maximum point of $f$,

which happens for equality case of two variables (when $z=y=u-\sqrt{u^2-v^2}$ in our case).

Since our inequality is homogeneous, we can assume $y=z=1$ and we need to prove that $$\sqrt{\frac{x^3+2}{x}}+\sqrt{\frac{2x+1}{x^2+2}}\geq1+\sqrt3$$ or $$\sqrt{\frac{x^3+2}{x}}-\sqrt3\geq1-\sqrt{\frac{2x+1}{x^2+2}}$$ or $$\frac{x^3-3x+2}{\sqrtx\left(\sqrt{(x^3+2)^2}+\sqrt{3x(x^3+2)}+\sqrt{9x^2}\right)}\geq\frac{x^2-2x+1}{\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)}$$ or $$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq\sqrtx\left(\sqrt{(x^3+2)^2}+\sqrt{3x(x^3+2)}+\sqrt{9x^2}\right).$$ Now, by C-S we obtain $$\sqrt{x^2+2}=\frac{1}{\sqrt3}\sqrt{(1+2)(x^2+2)}\geq\frac{x+2}{\sqrt3}$$ and $$\sqrt{2x+1}=\frac{1}{\sqrt3}\sqrt{(2+1)(2x+1)}\geq\frac{2\sqrt{x}+1}{\sqrt3}.$$ Also, by AM-GM $$3x\leq x^3+2.$$ Thus, it's enough to prove that $$(x+2)^2(x+2\sqrt{x}+3)\geq9\sqrt{x(x^3+2)^2}$$ or $$(a^2+2)^2(a^2+2a+3)\geq9\sqrt{a^2(a^6+2)^2},\tag1$$ where $a=\sqrt{x}$ and the rest is smooth.

The last inequality follows from the following inequalities. $$(a^2+2)^5\geq54a(a^6+2)$$ and $$2(a^2+2)(a^2+2a+3)^3\geq27a(a^6+2).$$ Done!

• I've learnt uvw method the first time from this problem. Thank you very much for your time. – carat Aug 10 '17 at 16:05
• @Hans It's not so easy, but I am ready to show. – Michael Rozenberg Aug 14 '17 at 7:59
• @Hans The second inequality is obvious. For the first: $(a^2+2)^5-54a(a^6+2)=(a^5-a^3-5a^2-4a+3)^2+(12a^2-44a+41)a^6+(6a^2-16a+11)a^4+(36a^2-34a+9)a^2+85a^2-84a+23>0$. – Michael Rozenberg Aug 15 '17 at 16:16
• @Hans If $a=b+1$ then $b$ can be negative. For the second inequality we have: $2(a^2+2)(a^2+2a+3)^3-27a(a^6+2)>2a^8-15a^7+46a^6>0$. Since the coefficient before $a$ in $2(a^2+2)(a^2+2a+3)^3$ greater than $54$, it's enough to calculate coefficients before $a^7$ and before $a^6$, which is easy. – Michael Rozenberg Aug 15 '17 at 18:10
• @Hans I tried, tried else and it turned out! Firstly, I want to delete $a^{10}$ and the rest is just playing with coefficients in $a^5+Aa^3+Ba^2+Ca+D$, – Michael Rozenberg Aug 15 '17 at 20:03

$$x+y+z=p,\ xy+yz+zx=q,\ xyz=r$$

we can assume : $x+y+z=1$

$x^3+y^3+z^3=1-3q+3r,\ x^2+y^2+z^2=1-2q\ \ \ \left( 0 < q \le \dfrac{1}{3}\right)$

$$\left( \dfrac{1-3q}{r}+3\right)^{1/3}+\left( \dfrac{q}{1-2q}\right)^{1/2}\ge 1+\sqrt3$$ We have $$q^2=(xy+yz+zx)^2\ge 3xyz(x+y+z) =3r.$$

So: $LHS \ge \sqrt3\cdot\left( \dfrac{1-3q}{q^2}+1\right)^{1/3}+\left( \dfrac{q}{1-2q}\right)^{1/2}=f(q)$

$$\dfrac{q}{1-2q}=t \le 1$$

$g(t)= \sqrt3\cdot\left( \dfrac{1+t}{t^2}-1\right)^{1/3}+ \sqrt{t}\ge g\left(1\right )=1+\sqrt3$

• I think a proof that $f$ decrease is not so easy. – Michael Rozenberg Aug 9 '17 at 9:54
• Yes, it is terrible function, but only one variable ))) – Sergic Primazon Aug 9 '17 at 10:11
• But indeed, it decreases. +1. – Michael Rozenberg Aug 9 '17 at 10:12
• Do you mean to say just $q^2\ge 3r$? Both $q\ge 3p$ and $pq=r$ are wrong. Take $x=y=z=\frac13$. – Hans Aug 9 '17 at 10:26
• @Hans It means $(xy+xz+yz)^2\geq3xyz(x+y+z)$. – Michael Rozenberg Aug 9 '17 at 10:27

A proof by using the Sergic Primazon's idea. Dedicated to dear Hans.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $$\sqrt{\frac{27u^3-27uv^2}{w^3}+3}+\sqrt{\frac{v^2}{3u^2-2v^2}}\geq\sqrt3+1.$$ Now, since $27u^3-27uv^2\geq0$ and $v^4\geq uw^3$, it's enough to prove that $$\sqrt{\frac{27u^4-27u^2v^2}{v^4}+3}+\sqrt{\frac{v^2}{3u^2-2v^2}}\geq\sqrt3+1.$$ Let $3u^2-2v^2=p^2v^2$, where $p>0$.

Thus, $p\geq1$, $u^2=\frac{p^2+2}{3}v^2$ and we need to prove that $$\sqrt{3(p^2+2)^2-9(p^2+2)+3}+\frac{1}{p}\geq1+\sqrt3$$ or $$\sqrt3\left(\sqrt{p^4+p^2-1}-1\right)\geq\frac{p-1}{p}$$ or $$\frac{\sqrt{3}(p^4+p^2-2)}{\sqrt{(p^4+p^2-1)^2}+\sqrt{p^4+p^2-1}+1}\geq\frac{p-1}{p}$$ or $$\sqrt3p(p+1)(p^2+2)\geq\sqrt{(p^4+p^2-1)^2}+\sqrt{p^4+p^2-1}+1$$ and since $$1\leq\sqrt{p^4+p^2-1},$$ it's enough to prove that $$\sqrt3p(p+1)(p^2+2)\geq3\sqrt{(p^4+p^2-1)^2}$$ or $$p^3(p+1)^3(p^2+2)^3\geq9(p^4+p^2-1)^2$$ or $$p^{12}+3p^{11}+9p^{10}+19p^9+21p^8+42p^7+26p^6+36p^5+33p^4+8p^3+18p^2-9\geq0.$$ Done!

• +1. Brilliant as usual! And thank you for your most generous dedication, Michael, hahahaa! – Hans Aug 10 '17 at 19:59
• @Hans You are welcome! – Michael Rozenberg Aug 10 '17 at 20:00
• Michael, could I possibly interest you in examining this inequality math.stackexchange.com/q/526050/64809? It is a tough one and has a bounty of 100. Thank you. – Hans Aug 10 '17 at 20:02
• @Hans I thought about this inequality. I still don't see a proof. – Michael Rozenberg Aug 10 '17 at 20:05
• OK. No worries, so long as you keep it simmering on your back burner. ;-) – Hans Aug 10 '17 at 20:06

Here is my, as Michael Rozenberg rightly dubs it, ugly proof of the last step in Sergic Primazon's proof that is $g(t)$ decreases for $t\in(0,1]$. $$g'(t) = -3^{-\frac23}\Big(\frac1{t^2}+\frac1t-1\Big)^{-\frac23}\Big(\frac2{t^3}+\frac1{t^2}\Big)+\frac12\Big(\frac1t\Big)^{\frac12}\ .$$ We want to show $g'(t)<0$ or $$3^{\frac23}\Big(\frac1t\Big)^{\frac12}\Big(\frac1{t^2}+\frac1t-1\Big)^{\frac23} < 2\Big(\frac2{t^3}+\frac1{t^2}\Big).$$

Let $s:=\frac1t-1\in[0,\infty)$. The above inequality is equivalent to it being raised to the power of $6$, because it is easy to see its both sides are positive after making the substitution of $t$ by $s$. The transformed inequality is $$2^6(s+1)^9(2s+3)^6-3^4((1+s)^2+s)^4>0 \tag1$$ The left hand side of the above inequality is $$4096 s^{15} + 73728 s^{14} + 617472 s^{13} + 3191808 s^{12} + 11388672 s^{11} + 29712384 s^{10} + 58555200 s^9 + 88764399 s^8 + 104358708 s^7 + 95156902 s^6 + 66743280 s^5 + 35363277 s^4 + 13701744 s^3 + 3665574 s^2 + 605556 s + 46575$$ All the coefficients of the polynomial are positive, so Inequality (1) holds for $s\ge 0$ implying $g'(t)<0$ for $t\in(0,1]$.

• It's terrible ugly, but it's proof. – Michael Rozenberg Aug 10 '17 at 21:08
• @MichaelRozenberg: Yes, it is ugly for sure. – Hans Aug 10 '17 at 21:09