With $x^2+y^2=1$ find Minimum and Maximum of $x^5+y^5$ (do not use derivative) It's easy to see that the minimum is $-1$ and maximum is $1$.
My idea is put $x=\cos(a)$, $y=\sin(a)$ and $t=x+y$,   so I have $-\sqrt{2}\le t \le \sqrt{2}$ then $0 \le t^2 \le 2$
When $t=x+y$ then $(x+y)^2=1+2xy$.
Then
\begin{aligned}
x^5+y^5&=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)\\
&=x^3+y^3-x^2y^2(x+y)\\
&=(x+y)(x^2-xy+y^2)-x^2y^2(x+y)\\
&=(x+y)(1-xy-x^2y^2)\\
&=\left( {x + y} \right)\left( {1+\frac{1}{2} - {{\left( {x + y} \right)}^2} - \frac{{{{\left( {x + y} \right)}^4}}}{4} + \frac{{{{\left( {x + y} \right)}^2}}}{2} - \frac{1}{4}} \right)\\
&=t\left( {\frac{5}{4}  - \frac{{{t^4}}}{4}} \right)\\
\end{aligned}
This problem is for those who have not studied derivatives so I dont have any idea for next step. Anyone can help me for the hint or other solutions? Very Thanks

If $x^2+y^2=1$, then $|x|$ and $|y|$ are each no larger than $1$. Consequently, $|x|^5\le x^2$ and $|y|^5\le y^2$. It follows that
$$|x^5+y^5|\le|x|^5+|y|^5\le x^2+y^2=1$$
