## ParthKohli one year ago Inequalities.

• This Question is Open
1. ParthKohli

If x, y, z are positive real numbers such that $$xyz = 32$$, then find the minimum value of $$x^2 + 4xy + 4y^2 + 2z^2$$.

2. ParthKohli

$= (x+2y)^2+2z^2\ge 2 \sqrt{(x+2y)(\sqrt2 z)}$

3. ParthKohli

oops.

4. ParthKohli

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

5. ParthKohli

6. freckles

7. ytrewqmiswi

loll XD

8. ParthKohli

hahaha

9. freckles

I was like halfway there

10. ParthKohli

so you removed all of it then?

11. freckles

yea

12. ParthKohli

well I mean you should have ended what you started

13. ytrewqmiswi

is it ok to assume x, y,z like x=2^a y=2^b z=2^c where a,b,c are some integers?

14. ParthKohli

Yes, but you can't assume that a, b, c are integers.

15. ytrewqmiswi

+ 1 more thing-> (a+b+c)=5 y can't we do so?

16. ParthKohli

yes, it's obvious that a+b+c=5 since 2^a 2^b 2^c = 2^5 just that we don't know if integers actually minimise the expression

17. ytrewqmiswi

ok then lets take a,b,c to be real numbers

18. ParthKohli

if that's the case, then I think we can complete the problem without taking the 2^x form. just that we still dunno what inequality to use.

19. ParthKohli

$\frac{(x+2y)^2 + 2z^2}{2} \ge \frac{2(x+2y)^2(2z)^2}{ (x+2y)^2 + 2z^2 }$not that it helps, but gotta try out all permutations of the am-gm-hm inequality haha

20. ytrewqmiswi

well a,b,c have to be real so substituting them in the equation we will have this- $2\times 2^a+4\times 2^a \times 2^b +2 \times 2^c$ using AM-GM $2^{2a}+2^{2+a+b}+2^{c+1} \ge 2 \sqrt {2^{2(a+b+c)+(a+b)+5}}$a+b+c=5$2^{2a}+2^{2+a+b}+2^{c+1} \ge 2\sqrt{2^{(a+b)+15}}$ so we need to find minimum of a+b since a and b are any real numbers we can take a+b=15 so $2^{2a}+2^{2+a+b}+2^{c+1} \ge 2\sqrt{2^{15-15}}$so minimum value occus at equality so min of$2^{2a}+2^{2+a+b}+2^{c+1} = 2$

21. ytrewqmiswi

wait we can get an even more minimum value by putting a+b=-(infty) 0-0 well this is not violating any of the given data??

22. ParthKohli

Here's the solution:$x^2+2xy+2xy+4y^2+z^2+z^2\ge 6\sqrt[6]{16x^4y^4z^4}=96$