If x+y is fixed prove x2+y2 is minimal when x = y

If $x + y = S > 0$ is fixed, prove that $x = y$ minimizes the sum $x^2 + y^2$ , where $x$ and $y$ are positive real numbers.

Proof. Since $x + y = S$ we have $y = S - x$ . Then, we want to minimize the function

$f(x) = x^2 + (S - x)^2.$

Taking the derivative we have

$f'(x) = 2x - 2 (S-x) = 4x - 2S.$

This means $f'(x) = 0$ when $x = \frac{S}{2}$ and we have

$\begin{align*} f'(x) &< 0 & \text{when} && x &< \frac{S}{2} \\ f'(x) &> 0 & \text{when} && x &> \frac{S}{2}. \end{align*}$

Therefore, $f$ is decreasing for $x < \frac{S}{2}$ and increasing for $x > \frac{S}{2}$ . Hence, $f(x)$ has a minimum at $x = \frac{S}{S}$ . This implies

$y = S - x = S - \frac{S}{2} = \frac{S}{2}.$

Hence, the sum is minimal when

$x = y. \qquad \blacksquare$

Stumbling Robot

If x+y is fixed prove x²+y² is minimal when x = y

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