If $x+y=2$ then show that $x^2y^2(x^2+y^2)≤2$

Question

Let $x$ and $y$ be two positive real numbers such that $x+y=2$. Then show that $$x^2y^2(x^2+y^2)≤2,$$

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In $x+y=2$ implies $x^3y^3(x^3+y^3)≤2$, for positive $x$ and $y$. we can see that $t=xy$ solves pretty simple. But here we get $$t^2(4-2t) \leq \Big({t+t+4-2t\over 3}\Big)^3 = {64\over 27}$$ which is greater than $2$.

Edit: Ahh, true it is not that much harder than I thought at first. Since $$0\leq t =xy\leq \Big({x+y\over 2}\Big)^2 =1$$ we have $$(t-1)(t^2-t-1)\geq 0$$ which is equivalent to $$t ^2(4-2t)\leq 2$$

Answer 1

Let us assume that $$x^2y^2(x^2+y^2)>2$$ A $x+y=2$, we have $x^2+y^2=4-2xy$. Substituting this in the inequality above, we get $$4x^2y^2-2x^3y^3>2$$ This implies that $$2x^2y^2>1+x^3y^3$$ This is a contradiction because $2x^2y^2\leq 1+x^4y^4$ by AM-GM inequality, and $1+x^4y^4\leq 1+x^3y^3$. The last inequality is true because $xy\leq 1$, which is something we get by applying the AM-GM inequality to the equation $x+y=2$.

Hence proved

Answer 2

it is equivalent to $$ (xy)^3-(xy)^2+\frac{1}{2}\geq 0$$ which is true. Consider the function $$h(t)=t^3-t^2+\frac{1}{2}$$ for $t>0$ $$h'(t)=3t^2-t$$ and $$h''(t)=6t>0$$ and $$h(1/3)>0$$

Answer 3

Put $x=2\cos^2(t) $

and $y=2\sin^2 (t) .$

then

$$x^2y^2=16\cos^4 (t)\sin^4 (t)=\sin^4 (2t)$$

and

$$x^2+y^2=4 \Bigl(1-2\sin^2 (t)\cos^2 (t)\Bigr)$$ $$=4-2\sin^2 (2t) $$ Let $f (z)=z^2 (2-z) $ for $0\le z\le 1$.

$$f'(z)=2z (2-z)-z^2=z (4-3z)$$

the maximum of $f (z) $ is attained for $z=1$ thus

$$x^2y^2 (x^2+y^2)=2f (\sin^2 (2t))\le 2f (1)\le 2$$

Answer 4

Let $x=1+k$ and $y=1-k$, for some $k\in(-1,1)$. Then \begin{align} x^2y^2(x^2+y^2)&=(1-k^2)^2\times2(1+k^2)\\ &=2(1-k^2)(1-k^4)\\ &\le 2 \end{align} Further, equality holds iff $k=0$, i.e. $x=y=1$.