Show that $x^2+y^2\geq x+y $.

Question

Let $x,y>0$ s.t. $x^3+y^3\geq 2$.

Show that $x^2+y^2\geq x+y $.

I analyse the case when $x,y\geq 1$ but I don't know to solve the case when $x\geq 1\geq y $.

Answer 1

Note that $$ \begin{align}(x+y)(x^2+y^2)&=x^3+y^3+xy(x+y)\\&\tag1\ge 2+xy(x+y)\\&=2+\frac{(x+y)^2-(x^2+y^2)}{2}(x+y)\\ &=2+\frac 12(x+y)^3-\frac12(x+y)(x^2+y^2)\end{align}$$ Therefore, $$ \tag2(x+y)(x^2+y^2)\ge \frac 43+\frac13(x+y)^3$$ So if the first factor is larger, we first find from $(1)$ $$ x+y>\sqrt2$$ and then from $(2)$ $$ x+y>\sqrt{\frac43+\frac13\sqrt 2^3},$$ which is larger. We can repeat this, i.e., if we know $x+y>a_n$, then also $x+y>\sqrt{\frac43+\frac13a_n^3}=:a_{n+1}$. The sequence $\{a_n\}_n$, starting with $a_1=\sqrt 2$, is strictly increasing. If it has a limit $a$, then $\sqrt{\frac43+\frac13a^3}=a$, i.e., $4+a^3-3a^2=0$, $a\in\{2,-1\}$. We conclude $$x+y\ge 2$$ As the case $x,y\ge1$ is trivial, we assume wlog $x=1+a>1$ and $y=1-b< 1$ with $a>b$. But then $$ x^2+y^2=1+2a+a^2+1-2b+b^2>2+2(a-b)>2+a-b=x+y.$$

Answer 2

By symmetry $x = y = z$

then

$$ 2z^3 \ge 2 \Rightarrow z^3 \ge 1 \Rightarrow z^2 \ge z $$

Comparison with $x^3+y^3 \ge 2$ (light blue) and the circle $(x-\frac{1}{2})^2+(y-\frac{1}{2})^2 = 1/2$ (red)

Answer 3

Consider $f(x,y)=x^2+y^2-x-y$. The inequality is equivalent to the fact that the following minimum is non-negative $$ \min f(x,y)\quad\text{subject to }x^3+y^3\ge 2,\ x\ge 0,\ y\ge 0. $$

1. As $f(x,y)=(x-\frac12)^2+(y-\frac12)^2-\frac12$ has compact sublevel sets, the minimum exists. Apply the necessary condition for optimality.
2. The first constraint must be active at the minimum (otherwise we can make the variable that is greater than one a bit smaller, which makes $f(x,y)$ smaller, contradiction). It is also easy to rule out the case when the positivity constraints are active ($x=0$ or $y=0$), the function $f(x,y)\ge 0$ there. Thus the only interesting case is where the gradients of $f(x,y)$ and $x^3+y^3$ are parallel. It makes $$ \begin{vmatrix}2x-1 & x^2\\2y-1 & y^2\end{vmatrix}=(x-y)(x+y-2xy)=0. $$
3. The case when $x+y=2xy$ is not interesting again since then $$ f(x,y)=x^2+y^2-2xy=(x-y)^2\ge 0. $$
4. The case $x=y$ and $x^3+y^3=2$ gives $x=y=1$ where $f(1,1)=0$. Hence, it is the minimum.