Prove an inequality $\frac{x^2}{1+y}+\frac{y^2}{1+x}\ge 1$ where $x,y \ge 0$ and $x+y=2$.

Question

I was trying to solve the following: $$ \frac{x^2}{1+y}+\frac{y^2}{1+x}\ge 1$$ $$x^2(1+x)+y^2(1+y)\ge (1+y)(1+x)$$ $$x^3+x^2+y^3+y^2 \ge 1+y+x+xy$$ $$(x+y)(x^2-xy+y^2)+x^2+y^2 \ge 1+y+x+xy$$ substitute $x+y=2$ $$2(x^2-xy+y^2)+x^2+y^2 \ge 3+xy$$ $$2((x+y)^2-3xy)+(x+y)^2-2xy \ge 3+xy$$ $$2(4-3xy)+4-2xy \ge 3+xy$$ $$12-8xy \ge 3+xy$$ $$1 \ge xy$$

Answer 1

And now you can use AM-GM: $$1=\frac{(x+y)^2}{4}\geq\frac{(2\sqrt{xy})^2}{4}= xy.$$

I think the following way a bit of better.

By C-S we obtain: $$\frac{x^2}{1+y}+\frac{y^2}{1+x}\geq\frac{(x+y)^2}{2+x+y}=1.$$ We can use also the Tangent Line method.

Indeed, $$\sum_{cyc}\frac{x^2}{1+y}-1=\sum_{cyc}\left(\frac{x^2}{3-x}-\frac{1}{2}\right)=$$ $$=\frac{1}{2}\sum_{cyc}\frac{2x^2+x-3}{3-x}=\frac{1}{2}\sum_{cyc}\frac{(x-1)(2x+3)}{3-x}=$$ $$=\frac{1}{2}\sum_{cyc}\left(\frac{(x-1)(2x+3)}{3-x}-\frac{5}{2}(x-1)\right) =\frac{9}{4}\sum_{cyc}\frac{(x-1)^2}{3-x}\geq0.$$

Answer 2

Use the fact, $$x=1+a$$ and $$y=1-a$$ where$0<a<1$. It will work, after you have substituted in the place of $x$ and $y$.respectively. Update 1: After substituting , we get, $$\dfrac{4+8a^2}{4-a^2}\geq 1 \Rightarrow \dfrac{9a^2}{4-a^2}\geq0$$ which is certainly true. And remember we can always consider $x=1+a$ and $y=1-a$ beacuse $x,y\in$R.

Answer 3

We have $$ \frac{x^2}{1+y}+\frac{y^2}{1+x}= \frac{x^2}{3-x}+\frac{y^2}{3-y}.$$ Put $f(x)=\frac{x^2}{3-x}.$ Since $$ f''(x)=\frac{18}{3-x} >0, $$ $f(x)$ is convex function for $x<3.$ Then by Jensen inequality $$ \frac{x^2}{3-x}+\frac{y^2}{3-y}=f(x)+f(y) \geq 2 f(\frac{x+y}{2})=2 f(1)=2 \cdot \frac{1}{2}=1. $$

Answer 4

Using Titu's lemma we get $$ \frac{x^2}{1+y}+\frac{y^2}{1+x}\ge \frac {{(x+y)}^2}{2+x+y}$$

Using the condition $x+y=2$ we get $$ \frac{x^2}{1+y}+\frac{y^2}{1+x}\ge \frac {2^2}{2+2}=1$$

Q. E. D