Question

Prove that :
\[\large{(x^3+y^3+z^3)^2 \ge \frac{9}{8}(x^2+yz)(y^2+zx)(z^2+xy)}\]

where x,y and z are non - negative real numbers.

Answer 1

should we expand here in RHS or in LHS ?


I did that in RHS but that was too much complicated ...result

Answer 2

@Callisto @myininaya @UnkleRhaukus

Answer 3

\[\text{if } \quad x=y=z=1\]
\[9\geq\frac98(8)\]

Answer 4

is it given that x = y = z = 1 ? @UnkleRhaukus

Answer 5

the equation is symmetric in the three variables

i think x=y=z=1
is a important point on the graph

Answer 6

Hmn but I don't think so as this as a correct way..

Answer 7

I am still waiting for @mukushla here...

He is good at inequalities..

Answer 8

try proving this
\[ \large{(x^3+y^3+z^3)^2 \ge \frac{3}{8}((x^2+yz)^3 +(y^2+zx)^3 + (z^2+xy)^3)}\]

Answer 9

not sure though ...

Answer 10

Hmn.... I will like to try that @experimentX . .... but please.... r u sure that this would help?

Answer 11

might ... I'm extremely busy due to exams!!

Answer 12

Oh k no problem.... best of luck for your exams....

Answer 13

@TuringTest and @KingGeorge @amistre64 might like to view this when they will be online

Answer 14

Here what I did so far which leads to the credibility that that the inequality is true:
If
\[
h(x,y,z)=\frac{\left(x^3+y^3+z^3\right)^2}{\left(x z+y^2\right)
\left(x^2+y z\right) \left(x y+z^2\right)}\\

\]
I computed its gradient. G and I verified that G(x,x,x)=0. Also
\[
h(x,x,x)=\frac 98
\]
One has to show that h has a minimum at those points and we will be done.
I am not there.

Answer 15

must say that I did the following and I hpe that I am very near to the answer

Answer 16

emm dont feorget u missing 3abc\[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)\]

Answer 17

and of course \[xy+yz+zx\le x^2+y^2+z^2\]

Answer 18

oh k I will try that again using your given hints.... @mukushla

Answer 19

\[27(x^{2}+yz)(y^{2}+zx)(z^{2}+xy)\leq (x^{2}+y^{2}+z^{2}+xy+yz+zx)^{3}\] \(\leq 8(x^{2}+y^{2}+z^{2})^{3}\)

Answer 20

But we need
\( \left( x^3+y^3+ z^3 \right)^2
\) and not \( \left( x^2+y^2+ z^2 \right)^3 \)
Am I missing something?

Answer 21

im stuck with that sir : to proving\[(x^{2}+y^{2}+z^{2})^{3} \le3 (x^{3}+y^{3}+z^{3})^{2}\]

Answer 22

i thought its obvious but now im stuck there

Answer 23

no problem @mukushla


actually @vishweshshrimali5 knew the correct answer and solution as he has just attempted that question today in olympiad and he says that he has a correct prooof.... He just challenged me... and I probably won the challenge... as I atleast tried it ..:P

as per my opiniion that is all about " expansion " .... but still there will be suspense

But today he is having rest I will let him know about this discussion and I am sure that he will declare the right proof here asap...

Answer 24

Proving the above inequality is the same as proving the original one.

Answer 25

A suggestion: The inequality's both sides are SYMMETRIC POLYNOMIALS in x,y,z And so they are expressible in products of lower order symmetric polynomials. Now lower order then 3 is either 2 or 1 and for these degrees a huge arsenal of harmonic, geometric and arithmetic MEAN inequalities is well known. (Mukusha probably knew the idea and he only showed the impressive consequence , not the internal principle)

Answer 26

power mean inequality\[ \sqrt[3]{\frac{x^{3}+y^{3}+z^{3}}{3}}\geq\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}} \]so\[(x^{2}+y^{2}+z^{2})^{3} \le3 (x^{3}+y^{3}+z^{3})^{2}\]

Answer 27

@mukushla Here is a challenging observation for an olympian: in every symmetric polynomial like here BOTH sides you can reduce the number of variables by 1 using uniform division by x^2y^2z^2 and intrducing 2 new ratio variables

Answer 28

thanks guys...

Answer 29

np :)