# Prove a geometric inequality, if 3 numbers are satisfying the condition

I just got this problem, but I have no idea on how to prove that.

Prove that if $x,y,z\in\mathbb{R},\ x,y,z\ge 0$ and $2\cdot(x\cdot z+x\cdot y+y\cdot z)+3\cdot x\cdot y\cdot z = 9$, then $(\sqrt x + \sqrt y + \sqrt z )^4 \ge 72$. This is a geometric inequality.

Can anyone help me, please? Any kind of help (solutions, hints etc) is really appreciated. Thank you!

NOTE: I REALLY DON'T KNOW WHAT TITLE SHOULD I WRITE FOR THIS POST, SO PLEASE LEAVE A COMMENT IF YOU HAVE AN IDEA FOR THE POST TITLE. THANK YOU.

• Two questions : 1) Where does this problem come from ? 2) What do you mean by "this is a geometric inequality" – Jean Marie Aug 10 '16 at 14:56
• @JeanMarie Thank you for your response. This problem comes from a spanish math book and they wrote "this is a geometric inequality". Thank you again. – MM PP Aug 11 '16 at 17:36
• The witty solution by @Michael Rozenberg is a feat. But the problem is that, faced to a similar issue another day, we won't necessarily be able to use the same "tricks". My attempt was to come back to the elementary symmetrical polynomials, which is a solid general base. – Jean Marie Aug 12 '16 at 4:58

Here is an attempt (I think in the good direction, thus a hint, as you wish it) to solve the problem, by putting it in a more tractable form with the elementary symmetrical polynomials in 3 variables, thanks to a change of variables.

Again, let us make it clear : it is not a solution (but I still work on it !).

Let us write the condition and the targetted constraint in order that the text is self-contained:

$$2(xz+xy+yz)+3xyz=9 \ \ \ (1)$$

and

$$(\sqrt x + \sqrt y + \sqrt z )^4 \ge 72 \ \ \ (2)$$

or by considering power 4 as two successive squarings:

$$(x+y+z+2(\sqrt{yz}+\sqrt{zx}+\sqrt{xy}))^2 \geq 72$$

Let us make the change of variables :

$$a=\sqrt{yz}, \ \ b=\sqrt{xz}, \ \ c=\sqrt{xy} \ \ (3)$$

the reciprocal formulas being (due to the assumed strict positivity of $x,y,z$):

$$x=\dfrac{bc}{a}, y=\dfrac{ca}{b}, z=\dfrac{ab}{c} \ \ \ (4)$$

Formula (1) becomes

$$2(a^2+b^2+c^2)+3abc=9 \ \ \ (1')$$

or

$$2(a+b+c)^2-4(ab+bc+ca)+3abc=9 \ \ \ (1'')$$

and constraint to be established becomes:

$$((x+y+z)+2(a+b+c))^2 \geq 72 \ \ \ (2')$$

But, using (4),

$$x+y+z=\dfrac{(bc)^2+(ca)^2+(ab)^2}{abc}=\dfrac{(bc+ca+ab)^2-2abc(a+b+c)}{abc} \ \ \ (5)$$

Plugging (5) in (3'), one gets, for the constraint to be established:

$$\left(\dfrac{(bc+ca+ab)^2}{abc}\right)^2 \geq 72 \ \ \ (2'')$$

Thus, setting

$$\begin{cases}s_1&=&a+b+c\\s_2&=&ab+ac+bc\\s_3&=&abc\\\end{cases}$$

Thus, the simplified problem is as follows :

Show that, under the condition (1'')

$$2s_1^2-4s_2+3s_3=9$$

(and knowing that all these quantities are positive) we are due to have (2''), i.e.,

$$s_2^4 \leq 72 s_3^2$$

Here, I am a little stucked... Newton identities (https://en.wikipedia.org/wiki/Newton%27s_identities) might may be of some help ?

We need to prove that $x+y+z\geq\sqrt[4]{72}$, where $x$, $y$ and $z$ are non-negatives such that $$\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=9$$ Let $x+y+z<\sqrt[4]{72}$, $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and $a+b+c=\sqrt[4]{72}$.

Hence, $k<1$ and $9=\sum\limits_{cyc}(2x^2y^2+x^2y^2z^2)=k^4\sum\limits_{cyc}(2a^2b^2+k^2a^2b^2c^2)<$

$<\sum\limits_{cyc}(2a^2b^2+a^2b^2c^2)$, which is a contradiction because we'll prove now that $$\sum\limits_{cyc}(2a^2b^2+a^2b^2c^2)\leq9$$ Indeed, we need to prove that $$9\left(\frac{a+b+c}{\sqrt[4]{72}}\right)^6\geq2(a^2b^2+a^2c^2+b^2c^2)\left(\frac{a+b+c}{\sqrt[4]{72}}\right)^2+3a^2b^2c^2$$ or $$(a+b+c)^6\geq16(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2+144\sqrt2a^2b^2c^2$$ Since by AM-GM

$(a+b+c)^4=(a^2+b^2+c^2+2(ab+ac+bc))^2\geq$

$\geq\left(2\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}\right)^2=8(a^2+b^2+c^2)(ab+ac+bc)$,

it remains to prove that $$(a+b+c)^2(a^2+b^2+c^2)(ab+ac+bc)\geq2(a^2b^2+a^2c^2+b^2c^2)(a+b+c)^2+18\sqrt2a^2b^2c^2$$ or $$(a+b+c)^2((a^2+b^2+c^2)(ab+ac+bc)-2(a^2b^2+a^2c^2+b^2c^2))\geq18\sqrt2a^2b^2c^2$$ or $$(a+b+c)^2\sum\limits_{cyc}(a^3b+a^3c-2a^2b^2+a^2bc)\geq18\sqrt2a^2b^2c^2$$ or $$(a+b+c)^2\sum\limits_{cyc}(ab(a-b)^2+a^2bc)\geq18\sqrt2a^2b^2c^2$$

Thus, it remains to prove that $(a+b+c)^3\geq18\sqrt2abc$, which is AM-GM:

$(a+b+c)^3\geq27abc\geq18\sqrt2abc$. Done!