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Prove an inequality of exponentials

by RoRi

For all x, y > 0 and for any constants a,b such that 0 < a < b prove that

    \[ \big( x^b + y^b \big)^{\frac{1}{b}} < \big( x^a + y^a \big)^{\frac{1}{a}}. \]

Proof. We want to consider the function

    \[ f(t) = (x^t + y^t)^{\frac{1}{t}}. \]

If we can show this function is decreasing on the positive real axis then we establish the inequality since this would mean that if 0 < a < b then

    \[ f(b) < f(a) \quad \implies \quad (x^b+y^b)^{\frac{1}{b}} < (x^a+y^a)^{\frac{1}{a}}.\]

(So, the trick here is to think of this as a function of the exponent. The x and y are some positive fixed constants.) To take the derivative of f(t) we use logarithmic differentiation,

    \begin{align*} &&f(t) &= (x^t+y^t)^{\frac{1}{t}} \\[10pt] \implies &&\log f(t) &= \frac{1}{t} \log (x^t+y^t) \\[10pt] \implies &&\frac{d}{dt} (\log f(t)) &= \frac{d}{dt} \left( \frac{1}{t} \log (x^t+y^t) \right) \\[10pt] \implies &&\frac{f'(t)}{f(t)} &= -\frac{1}{t^2} \log (x^t+y^t) + \frac{1}{t} \left( \frac{1}{x^t+y^t} \right) \left( x^t \log x + y^t \log y \right) \\[10pt] \implies &&\frac{f'(t)}{f(t)} &= \frac{-\log(x^t+y^t)}{t^2} + \frac{x^t \log x + y^t \log y}{t (x^t+y^t)}. \end{align*}

Multiplying both sides by f(t) we then obtain

    \begin{align*} f'(t) &= \left( \frac{-\log(x^t+y^t)}{t^2} + \frac{x^t \log x + y^t \log y}{t(x^t+y^t)} \right) (x^t+y^t)^{\frac{1}{t}} \\[10pt] &= (x^t + y^t)^{\frac{1}{t}} \left( \frac{t(x^t \log x + y^t \log y) - (x^t+y^t)\log(x^t+y^t)}{t^2 (x^t+y^t)} \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t} - 1}}{t^2} \right) \left( x^t t \log x + y^t t \log y - (x^t + y^t)\log(x^t+y^t) \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t}-1}}{t^2} \right) \left( x^t \log x^t + y^t \log y^t - x^t \log (x^t+y^t) - y^t \log (x^t+y^t) \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t} - 1}}{t^2} \right) \left( x^t \log \left(\frac{x^t}{x^t+y^t}\right) + y^t \log \left( \frac{y^t}{x^t+y^t} \right) \right). \end{align*}

Now we can conclude that f'(t) < 0 for all t > 0 since the first term in the product

    \[ \frac{(x^t+y^t)^{\frac{1}{t}-1}}{t^2} > 0. \]

Since x, y > 0 (any real power of a positive number is still positive) and t^2 > 0. For the second term we have

    \[ x^t \log \left( \frac{x^t}{x^t+y^t} \right) + y^t \log \left( \frac{y^t}{x^t+y^t} \right) < 0 \]

since x^t and y^t are positive, but both logarithms are negative. We know these logarithms are negative since

    \[ \frac{x^t}{x^t+y^t} < 1 \quad \text{and} \quad \frac{y^t}{x^t+y^t} < 1 \]

implies

    \[ \log \left( \frac{x^t}{x^t+y^t} \right) < 0 \quad \text{and} \quad \log \left( \frac{y^t}{x^t+y^t} \right) < 0. \]

Hence, f'(t) < 0 for all t > 0. This means f(t) is a decreasing function. Therefore, if 0 < a < b then we have

    \[ f(b) < f(a) \quad \implies \quad \big( x^b + y^b \big)^{\frac{1}{b}} < \big( x^a + y^a \big)^{\frac{1}{a}}. \qquad \blacksquare \]

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