Suppose $a= f(t)>=0$ and $b= g(t) >= 0$ . Can we say that maximum value of $f(t)g(t)$ is $\frac{(f(t) + g(t))^2}{4}$ using AM-GM ? So we can just find maximum of function $\frac{[f(t)+g(t)]^2}{4}$ and lets say its equal to $M$ then $M>= f(t)g(t)$ ? Max Achieved on solving $f(t)=g(t)$? Or is there some condition imposed on f(t) and g(t) as then only this method will work ?
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Have you tried this with specific nonconstant functions $f(t)$ and $g(t)$? Counterexamples are easy to find. – Greg Martin Mar 17 '22 at 22:45
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@GregMartin yes i know it can fail , i mean i wanted to know when this doesnt fails the conditions i mean on f(t) and g(t) – Orion_Pax Mar 18 '22 at 12:13
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In that case, please edit your post to clarify the question you are asking and to remove the questions that you already know the answers to. – Greg Martin Mar 18 '22 at 16:09