3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Two questions : 1) Where does this problem come from ? 2) What do you mean by "this is a geometric inequality" $\endgroup$ – Jean Marie Aug 10 '16 at 14:56
  • $\begingroup$ @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. $\endgroup$ – MM PP Aug 11 '16 at 17:36
  • $\begingroup$ 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. $\endgroup$ – Jean Marie Aug 12 '16 at 4:58
1
$\begingroup$

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 ?

$\endgroup$
1
$\begingroup$

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!

$\endgroup$

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.