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Hi miwalin!

First, let's agree that $x,y\geq 0$ for the definitions to make sense.

The solution is then actually quite simple: Let's assume $x^{2y}\geq y^{2x}\geq 0$. Then $x^{2y}+y^{2x}\leq 1$ is equivalent to $x^{4y}-y^{4x}\leq x^{2y}-y^{2x}$, i.e. $x^{2y}(x^{2y}-1)\leq y^{2x}(y^{2x}-1)$. By assumption, it suffices to show $0\geq x^{2y}-1\geq y^{2x}-1$, that is $x^{2y}\leq 1$ on the set $x+y=1$.

EDIT: next paragraph is corrected. I just computed the derivative wrong .

But this is standard: For instance, observe that $f(x):=x^{2-2x}$ takes values $f(0)=0, f(1)=1$ and its derivative $x^{2-2x}(\frac{2-2x}{x}-2\ln(x))$ is $\geq 0$, because $x^{2-2x}\geq 0$ and $\frac{1-x}{x}\geq \ln(x)$ for $x\in (0,1]$.

I hope this helped.proof attempt
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Hi miwalin!

First, let's agree that $x,y\geq 0$ for the definitions to make sense.

The solution is then actually quite simple: Let's assume $x^{2y}\geq y^{2x}\geq 0$. Then $x^{2y}+y^{2x}\leq 1$ is equivalent to $x^{4y}-y^{4x}\leq x^{2y}-y^{2x}$, i.e. $x^{2y}(x^{2y}-1)\leq y^{2x}(y^{2x}-1)$. By assumption, it suffices to show $0\geq x^{2y}-1\geq y^{2x}-1$, that is $x^{2y}\leq 1$ on the set $x+y=1$.

EDIT: next paragraph is corrected. I seem to have just computed the derivative wrong.

But this is standard: For instance, observe that $f(x):=x^{2-2x}$ takes values $f(0)=0, f(1)=1$ and its derivative $x^{2-2x}(\frac{2-2x}{x}-2\ln(x))$ is $\geq 0$, because $x^{2-2x}\geq 0$ and $\frac{1-x}{x}\geq \ln(x)$.ln(x)$ for $x\in (0,1]$.

I hope this helped.
show/hide this revision's text 2 added 85 characters in body

Hi miwalin!

First, let's agree that $x,y\geq 0$ for the definitions to make sense.

The solution is then actually quite simple: Let's assume $x^{2y}\geq y^{2x}\geq 0$. Then $x^{2y}+y^{2x}\leq 1$ is equivalent to $x^{4y}-y^{4x}\leq x^{2y}-y^{2x}$, i.e. $x^{2y}(x^{2y}-1)\leq y^{2x}(y^{2x}-1)$. By assumption, it suffices to show $0\geq x^{2y}-1\geq y^{2x}-1$, that is $x^{2y}\leq 1$ on the set $x+y=1$.

EDIT: next paragraph is corrected. I seem to have computed the derivative wrong.

But this is standard: For instance, observe that $f(x):=x^{2-2x}$ takes values $f(0)=0, f(1)=1$ and its derivative $x^{2-2x}\cdot\frac{4x-4}{x}\ln(x)$ x^{2-2x}(\frac{2-2x}{x}-2\ln(x))$ is $\geq 0$, because $\ln(x)/x\leq 0, 4x-4\leq x^{2-2x}\geq 0$ for and $x\in [0,1]$.\frac{1-x}{x}\geq \ln(x)$.

I hope this helped.
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