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Prove that, for $1 \leq p<\infty$ and $a, b \geq 0$, $$ (a+b)^p=\inf _{0<t<1} t^{1-p} a^p+(1-t)^{1-p} b^p . $$ I tried to solve the problem as follows.

The assertion is clearly true if $p=1$ or $a b=0$. Hence we may suppose that $1<p<\infty$ and $a, b>0$. The smooth function $f(t):=$ $t^{1-p} a^p+(1-t)^{1-p} b^p$ for $t \in(0,1)$ satisfies $\lim _{t \to 0^+} f(t)=+\infty=\lim _{t \to 1^-} f(t)$. Hence $f$ has a minimum somewhere in $(0,1)$. To find it we let $f^{\prime}(t)=0$ and we obtain: $$ f'(t) = (1-p) t^{-p} a^p - (1-p) (1-t)^{-p} b^p. $$ i.e. $$ t^{-p} a^p = (1-t)^{-p} b^p. $$ I'm stuck on these accounts, how can I proceed?

Edit Last steps. From $$\left(\frac{1-t}{t}\right)^p=\left(\frac{b}{a}\right)^p$$ I obtain $$\frac{1}{t}-1=\frac{b}{a}\quad \Rightarrow \quad \frac{1}{t}=1+\frac{b}{a}$$ and then $$t=\frac{a}{a+b}$$.

Lely
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    You are almost there, you just have to solve for $t$, then plug the value of $t$ in $f(t)$ and the proof is done. If needed I can post a solution from where you got stuck, but you are almost done so I feel you would be happy to finish on your own. – SotArmen Jul 17 '23 at 10:43
  • @SotArmen maybe I succeeded – Lely Jul 18 '23 at 06:42
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    Yes, you have succeded, if you substitute $t$ into $f(t)$ , you get the left hand side. – SotArmen Jul 18 '23 at 12:59

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