Prove that $\dfrac {2x+y+z}{x}+\dfrac {2y+x+z}{y}+\dfrac {2z+x+y}{z}\ge 12$

Question

I am trying to prove that

$$\dfrac {2x+y+z}{x}+\dfrac {2y+x+z}{y}+\dfrac {2z+x+y}{z}\ge 12$$

Clearly

$2x+y+z>x, \ 2y+x+z>y, 2z+x+y>z$

However, I struggle to find a suitable inequality relation to move further on here. Does anyone have a suggestion?

Thanks

You should tell us where $x,y,z$ live--are they real numbers? Also, please show any work you've initially done. — Nick F, Commented Aug 21, 2023 at 12:35
@NickF Now we can re-write this to $$\dfrac{(x^2 y + x^2 z + x y^2 + 6 x y z + x z^2 + y^2 z + y z^2)}{x y z}\geq 12$$ and since we have the following equality: $$(x+y+z)(1/x+1/y+1/z)=\frac{(x + y + z) (x y + x z + y z)}{x y z}$$ We can further develop it. But how do we go on from there? — Superunknown, Commented Aug 21, 2023 at 12:51

kleinbottle · Accepted Answer · 2023-08-21 12:43:31Z

4

LHS = $3 + (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \ge 3 + 9 = 12$ by AM-HM inequality.

answered Aug 21, 2023 at 12:43

kleinbottle

8416 silver badges15 bronze badges

$\begingroup$ How did you get that? $\endgroup$
– Superunknown
Commented Aug 21, 2023 at 12:47
$\begingroup$ Get what? The first line or the inequality? $\endgroup$
– kleinbottle
Commented Aug 21, 2023 at 12:50
1

$\begingroup$ en.wikipedia.org/wiki/HM-GM-AM-QM_inequalities $\endgroup$
– kleinbottle
Commented Aug 21, 2023 at 12:53
1

$\begingroup$ $\frac{2x+y+z}{x} = 1 + \frac{x+y+z}{x}$. Can you take it from there? $\endgroup$
– kleinbottle
Commented Aug 21, 2023 at 13:02
1

$\begingroup$ No, do the same thing to each of the three terms, that's where the 3 comes from and you are left with $(x+y+z)(1/x+1/y+1/z)$. $\endgroup$
– kleinbottle
Commented Aug 21, 2023 at 13:12

| Show 2 more comments

PRANSHU GOYAL · Accepted Answer · 2023-08-21 16:14:29Z

2

SEE (2x+y+z)/x = (x+y+z+x)/x = {(x+y+z)/x}+1 similarly do this for other numbers you will obtain

(x+y+z)(1/x + 1/y + 1/z) + 3 by am-hm inequality this is greater than equal to 12 Hence Proved

answered Aug 21, 2023 at 16:14

PRANSHU GOYAL

362 bronze badges

$\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$
– José Carlos Santos
Commented Aug 21, 2023 at 16:40
$\begingroup$ How do you actually show that $(x+y+z)(1/x + 1/y + 1/z) + 3\geq 12$ here? $\endgroup$
– Superunknown
Commented Sep 5, 2023 at 11:30

Add a comment |

2 Answers 2