Proove that for real numbers p ang q $\displaystyle \therefore$ $\displaystyle \frac{1}{p}+\frac{1}{q}=1$ ,

$\displaystyle (2p)^{2013}+(2q)^{2013}>2(pq)^{2013}$

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- Feb 6th 2013, 09:09 AMdarence(2p)^2013+(2q)^2013>2(pq)^2013
Proove that for real numbers p ang q $\displaystyle \therefore$ $\displaystyle \frac{1}{p}+\frac{1}{q}=1$ ,

$\displaystyle (2p)^{2013}+(2q)^{2013}>2(pq)^{2013}$ - Feb 6th 2013, 10:25 AMebainesRe: (2p)^2013+(2q)^2013>2(pq)^2013
It's not true for all real numbers p and q. For example if p = 1/2 and q = -1 the left hand side is less than the right hand side for odd values of the exponent.

I suspect the problem should have been written "for__positive__real values of p and q..." In this case you can show that p = q/(q-1) and that p and q are both > 1. The equation boils down to $\displaystyle (q-1)^n > -1 }$, where 'n' is the exponent (2013 in this case). - Feb 6th 2013, 11:08 AMdarenceRe: (2p)^2013+(2q)^2013>2(pq)^2013
Omg:(

This was on my examination and I don't pass because I could not proove.

Thank you very much. - Feb 6th 2013, 10:51 PMearthboyRe: (2p)^2013+(2q)^2013>2(pq)^2013
like ebaines, I guess the real question was: prove that for positive reals p and q, $\displaystyle \frac{1}{p}+\frac{1}{q}=1$ ,

$\displaystyle (2p)^{2013}+(2q)^{2013}>2(pq)^{2013}$

then, I think its pretty obvious because $\displaystyle \frac{1}{p}+\frac{1}{q}=1 \implies p+q=pq$ (now for the given relation to hold remember that p and q must be both positive and also greater than 1)

$\displaystyle p+q=pq \implies (p+q)^{2013}=pq^{2013} \implies p^{2013}+q^{2013}+\underbrace{........}_{\mathclap {\text{other positive terms of expansion}}}=(pq)^{2013}$

$\displaystyle \implies p^{2013}+q^{2013}>(pq)^{2013} \implies 2^{2013}(p^{2013}+q^{2013})>2^{2013}(pq)^{2013}>2( pq)^{2013}$

hence $\displaystyle (2p)^{2013}+(2q)^{2013}>2(pq)^{2013}$ - Feb 7th 2013, 03:48 AMdarenceRe: (2p)^2013+(2q)^2013>2(pq)^2013
It's OK, Thank you very much. But problem was as I wrote and I failed exam :( I will see what to do now...