# Solving an inequality

• Sep 12th 2009, 09:29 AM
BG5965
Solving an inequality
Let x and y be positive real numbers. Prove that:

$(p + 2)(q + 2)(p + q) \geq 16pq$

I'm not sure how to approach this. Do I simplify/expand somehow or is there a special thing I should do with problems like these.

• Sep 12th 2009, 11:08 AM
red_dog
Use AM-GM inequality:

$p+2\geq 2\sqrt{2p}$

$q+2\geq 2\sqrt{2q}$

$p+q\geq 2\sqrt{pq}$

Now multiply the inequalities.
• Sep 13th 2009, 12:19 AM
dhiab
Quote:

Originally Posted by BG5965
Let x and y be positive real numbers. Prove that:

$(p + 2)(q + 2)(p + q) \geq 16pq$

I'm not sure how to approach this. Do I simplify/expand somehow or is there a special thing I should do with problems like these.

Hello there is the solution:
I'have :

$\left( {p + 2} \right)\left( {q + 2} \right)\left( {p + q} \right) = p^2 q + q^2 p + 4\left( {p + q} \right) + 2\left( {p^2 + q^2 } \right) + 4pq
$

but :

$\frac{{p^2 q + q^2 p + 4p + 4q}}{4} \ge \sqrt[4]{{16p^4
q^4 }} = 2pq
$

$p^2 q + q^2 p + 4p + 4q \ge 8pq.....(1)$

$\frac{{2p^2 + 2q^2 }}{2} \ge \sqrt {4p^2 q^2 } = 2pq$

$2\left( {p^2 + q^2 } \right) \ge 4pq.....(2)$

$(1) + (2):p^2 q + q^2 p + 4\left( {p + q} \right) + 2\left( {p^2 + q^2 } \right) \ge 8pq + 4pq$
Conclusion :

$
p^2 q + q^2 p + 4\left( {p + q} \right) + 2\left( {p^2 + q^2 } \right) + 4pq \ge 8pq + 4pq + 4pq = 16pq$
• Sep 15th 2009, 09:08 AM
BG5965
Sorry, dhiab, but I don't 100% understand your post - especially everything after step 1 (the expansion). E.g. why is $16pq$ suddenly $\sqrt[4]{16p^4 q^4}$ and everything after that.

Could you please explain further. Thanks, BG