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

$x^2+y^2=1$ implies $x \in [-1,1]$ and $y \in [-1,1]$
(You can verify this by the fact that all (x,y) lie on the unit circle of radius 1 so x and y must lie in the interval [-1,1])
Now, $x^2 \in [0,1]$ and $y^2 \in [0,1]$
Also $x^3 \in [-1,1]$ (since $x \in [-1,1]$)
and $y^3 \in [-1,1]$ (since $y \in [-1,1]$)
Now $$-1 \leq x^3 \leq 1 \implies -x^2 \leq x^5 \leq x^2 \tag{i}$$ (multiplied $x^2\geq0$ throughout the inequality)
Similarly $$-1 \leq y^3 \leq 1 \implies -y^2 \leq y^5 \leq y^2 \tag{ii}$$ (multiplied $y^2\geq0$ throughout the inequality)
From inequalities (i) and (ii)
$$-x^2-y^2 \leq x^5+y^5 \leq x^2+y^2$$
$$ -1\leq x^5+y^5 \leq 1$$
Since $x^5+y^5$ actually takes values $-1$ and $1$ for atleast one $(x,y)$ such that $x^2+y^2=1$ (for instance $x^5+y^5=-1$ for $x=-1, y=0$ and $x^5+y^5=1$ for $x=1, y=0$)
Therefore -1 and 1 are minimum and maximum values of $x^5+y^5$ respectively.
