Murad.Aghazade wrote: $\boxed{\text{A4}}$ Let $x,y$ be positive reals satisfying the condition $x^3+y^3\leq x^2+y^2$.Find the maximum value of $xy$. Method 1: \[xy \leq \frac{1}{4}(x+y)^2 \leq \frac{(x^3+y^3)^2}{(x^2+y^2)^2} \leq \frac{(x^2+y^2)^2}{(x^2+y^2)^2}=1\]Equality for $x=y=1$

My solution by Chebishev we have $x^2+y^2\geq x^3+y^3\geq \frac{(x+y)(x^2+y^2)}{2}\geq (xy)^{\frac{1}{2}}\cdot (x^2+y^2)$ A.M-G.M which means $xy\leq 1$

Murad.Aghazade wrote: $\boxed{\text{A4}}$ Let $x,y$ be positive reals satisfying the condition $x^3+y^3\leq x^2+y^2$.Find the maximum value of $xy$. Method 2: Let $y=kx$, then we have: $x \leq \frac{k^2+1}{k^3+1}$, and: \[xy=kx^2 \leq k\frac{(k^2+1)^2}{(k^3+1)^2}=1-\frac{(k-1)^2(k^4+k^3+k^2+k+1)}{(k+1)^2(k^2-k+1)^2} \leq 1\]

Murad.Aghazade wrote: $\boxed{\text{A4}}$ Let $x,y$ be positive reals satisfying the condition $x^3+y^3\leq x^2+y^2$.Find the maximum value of $xy$. Method 3: We can use the following inequality: $ln(x) \leq x-1 \leq x^3-x^2$ which gives: \[ ln(xy)=ln(x)+ln(y) \leq x^3-x^2+y^3-y^2 \leq 0\]Then we conclude that: $xy \leq 1$.

Murad.Aghazade wrote: $\boxed{\text{A4}}$ Let $x,y$ be positive reals satisfying the condition $x^3+y^3\leq x^2+y^2$.Find the maximum value of $xy$. Method 4: if both $x$ and $y$ $ \leq 1$, the problem is obvious. Without loss of generality, we can assume that $x \leq y$, then $x \leq 1 \leq y$. Let $f(t) = t^3-t^2+x^3-x^2$, then $f$ is strictly increasing for $t \geq 1$ and we have $f(y) \leq 0$. If we assume that $y > \frac{1}{x}$, than $f(y) > f(\frac{1}{x}) =\frac{(x-1)^2(x^4+x^3+x^2+x+1)}{x^3} \geq 0$ Contradiction! then: $y \leq \frac{1}{x}$ which means that $xy \leq 1$.

Murad.Aghazade wrote: $\boxed{\text{A4}}$ Let $x,y$ be positive reals satisfying the condition $x^3+y^3\leq x^2+y^2$.Find the maximum value of $xy$. Method 5 Using Cauchy-Schawrtz, we can prove that: \[x^3+y^3 \leq x^2+y^2 \leq x+y \leq 2\]Then: $xy \leq \frac{x^2+y^2}{2} \leq 1$. Equality for $x=y=1$.

Murad.Aghazade wrote: $\boxed{\text{A4}}$ Let $x,y$ be positive reals satisfying the condition $x^3+y^3\leq x^2+y^2$.Find the maximum value of $xy$. Method 6: We can use Power Mean: \[\sqrt[2]{\frac{x^2+y^2}{2}} \leq \sqrt[3]{\frac{x^3+y^3}{2}} \leq \sqrt[3]{\frac{x^2+y^2}{2}}\]Then we conclude that $x^2+y^2 \leq 2$ and by AM-GM \[xy\leq \frac{x^2+y^2}{2} \leq 1\]Equality for $x=y=1$.

bel.jad5 wrote: Murad.Aghazade wrote: $\boxed{\text{A4}}$ Let $x,y$ be positive reals satisfying the condition $x^3+y^3\leq x^2+y^2$.Find the maximum value of $xy$. Method 5 Using Cauchy-Schawrtz, we can prove that: \[x^3+y^3 \leq x^2+y^2 \leq x+y \leq 2\]Then: $xy \leq \frac{x^2+y^2}{2} \leq 1$. Equality for $x=y=1$. hey can u pls elaborate this solution