It is clear if $a \geq 1$ or $b \geq 1$, but how can I show it when $0< a,b <1$?
I want hints.
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2Maybe try Bernoulli’s inequality. – Sahaj Feb 27 '24 at 18:42
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2Bernoulli is a good start. As a hint for how this ends (which requires intermediate work), show that $ a^ b \geq a/(a+b)$, hence the result follows. – Calvin Lin Feb 27 '24 at 19:27
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@Sahaj thank you, I will try. – Mahmoud albahar Feb 27 '24 at 19:40
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@CalvinLin I will do the intermediate work, thank you, the beauty of learning lies in trying and putting efforts not getting the solution without trying. – Mahmoud albahar Feb 27 '24 at 19:41
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Also: https://math.stackexchange.com/q/381090/42969, https://math.stackexchange.com/q/319578/42969, https://math.stackexchange.com/q/1977318/42969 – all found with Approach0 – Martin R Feb 27 '24 at 19:53
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If $0 <a<1$ and $0 < b < 1$ then $\frac1b > 1$ then $$b^{a-1} = \left(\frac1b\right)^{1-a} \ge 1 \ge 1-a$$ this implies $$b^a \ge b(1-a)$$ Finally, $$a^b + b^a \ge (1-(1-a))^b + b(1-a) \ge 1$$
Kroki
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