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I'm trying to prove that $\forall a\in\mathbb R$, the value of $x$ that gives the maximum value of $y=x(a-x)$ is $x=\frac{a}{2}$.

I'm told that I must use this inequality to prove this: $\forall x,y\in\mathbb R$, $xy\leq(\frac{x+y}{2})^2$.

I know how to do this with basic calculus by finding $\frac{dy}{dx}=a-2x$ then $\frac{dy}{dx}=0$ when $x=\frac{a}{2}$.

I have no clue how to even start with using the above inequality to prove this though... Any help would be greatly appreciated!

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  • $\begingroup$ Alternatively you can complete the square (which is also a proof of the inequality that they told you to use) $x(a-x)=a^2/4-(x-a/2)^2$. $\endgroup$ Jan 26, 2020 at 23:03

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Nevermind, figured it out minutes after posting this.

$$\begin{align*} x(a-x)&\leq(\frac{x+a-x}{2})^2\\ y&\leq(\frac{a}{2})^2\\ \end{align*}$$ Hence the max of $y$ is $\frac{a^2}{4}$. Solving for $x$: $$\begin{align*} \frac{a^2}{4}&=ax-x^2\\ x^2-ax+\frac{a^2}{4}&=0 \end{align*}$$ This has a double root when $x=\frac{a}{2}$.

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