Prove that (a^x+b^x)^(1/x) > (a^y+b^y)^(1/y) for a > 0, b

Prove that $\bigl(a^{x}+b^{x}\bigr)^{1/x} \gt \bigl(a^{y}+b^{y}\bigr)^{1/y}$ for $a\gt 0$, $b\gt 0$ and $0\lt x\lt y$.

Step 1: Reduce to a monotonicity statement

Take logarithms and set

\[ F(t) = \frac{1}{t}\ln\bigl(a^{t}+b^{t}\bigr), \qquad t\gt 0. \]

The claim is exactly that $F$ is strictly decreasing on $\bigl(0,\infty\bigr)$.

Step 2: Differentiate

With $S(t) = a^{t}+b^{t}$ and $S'(t) = a^{t}\ln a+b^{t}\ln b$, $F'(t) = \frac{\dfrac{t\,S'(t)}{S(t)}-\ln S(t)}{t^{2}}$. so the sign of $F'$ is the sign of

\[ N(t) = \frac{t\,S'(t)}{S(t)}-\ln S(t). \tag{1} \]

Step 3: A change of variable

Put $u = a^{t}\gt 0$ and $v = b^{t}\gt 0$, so that $\ln u = t\ln a$ and $\ln v = t\ln b$; then $t\,S'(t) = u\ln u+v\ln v$ and (1) becomes

\[ N = \frac{u\ln u+v\ln v}{u+v}-\ln\bigl(u+v\bigr). \tag{2} \]

So $F'\lt 0$ is equivalent to the inequality

\[ u\ln u+v\ln v \lt \bigl(u+v\bigr)\ln\bigl(u+v\bigr) \qquad (u,v\gt 0). \tag{3} \]

Step 4: Proof of (3)

Fix $v\gt 0$ and set $h(u) = \bigl(u+v\bigr)\ln\bigl(u+v\bigr)-u\ln u-v\ln v$. Then

\[ h'(u) = \ln\bigl(u+v\bigr)+1-\ln u-1 = \ln\left(1+\frac{v}{u}\right) \gt 0, \] so $h$ is strictly increasing in $u$; and $h(u)\to v\ln v-v\ln v = 0$ as $u\to0^{+}$, using $u\ln u\to0$ (Problem 391). Hence $h(u)\gt 0$ for all $u\gt 0$, which is (3). Note that nothing here requires $a\neq b$.

Step 5: Conclusion

Thus $N(t)\lt 0$ and $F'(t)\lt 0$ for every $t\gt 0$, so $F$ is strictly decreasing. For $0\lt x\lt y$ this gives $F(x)\gt F(y)$, and exponentiating restores the assertion.

Step 6: Verification

\[a = 2,\ b = 3:\quad t = 0.5:\ 9.8989795; \quad t = 1:\ 5; \quad t = 2:\ 3.6055513 \;\;\checkmark\]

\[t = 4:\ 3.1382890; \qquad t = 10:\ 3.0051623 \;\;\checkmark\ \text{(decreasing towards } \max(a,b) = 3)\]

\[a = b:\ F(t) = \frac{\ln2}{t}+\ln a,\ \text{so } F'(t) = -\frac{\ln2}{t^{2}}\lt 0 \;\;\checkmark\]