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given : $x>0 ,y >0 ,b>a>0$
prove the following by using derivative of a appropriate function:
$${(x^b+y^b)}^{(1/b)} < {(x^a+y^a)}^{(1/a)}$$

I tried using $f(x)=(m^x+n^x)^{(1/x)}$ and
$f(x)={(1+k^x)}^{(1/x)}$

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user143997
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  • $f(x) = (1 + k^x)^{1/x}$ seems reasonable. Did you try computing the derivative? What did you get? – Aryabhata Jan 09 '15 at 02:35
  • thank you guys ,i already figured it out let f(x)=(1+kx)^(1/x). and let g(x) = lnf(x).computer the derivative of g(x) , g^(1)(x) = {x(k^x)/(1+k^x)}logk - ln(1 + k^x) = (k^x)/(1+k^x)ln(k^x)-ln(1 + k^x) < ln(k^x)-ln(1 + k^x) <ln(1 + k^x) -ln(1 + k^x) = 0 – user143997 Jan 09 '15 at 03:45

1 Answers1

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$f(t) = (x^t+y^t)^{1/t} \to \ln (f(t)) = \dfrac{\ln(x^t+y^t)}{t} \to \dfrac{f'(t)}{f(t)} = \dfrac{\dfrac{\ln x^t\cdot x^t+\ln y^t\cdot y^t}{x^t+y^t}-\ln(x^t+y^t)}{t^2}$. Substitute $u = x^t, v = y^t\to u\ln u + v\ln v <(u+v)(\ln u + \ln v)$ which is true if $u, v > 1$ and $f$ decreases, so $f(b) < f(a)$. In doing this we assume further $b, a > 1$.

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