I ran into this problem a few days ago, and have been trying to prove it, but all my attempts have been unsuccessful.So I am asking for your help. Here's the problem:
If $xy=(x+y)^2-5(x+y)+8$ for some $x,y\in\mathbb{R}$ then prove the following inequality holds $x^2+y^2\leq8$.
On
hint: Use Lagrange Multiplier !. I think LM does a very good job with this particular problem. It bypasses the cumbersome part of the work.
On
$$x^2 + x(y -5)-5y +y^2+ 8 = 0$$
For this to have a real solution $x$, we have
$$b^2-4ac = (y-5)^2 - 4(y^2-5y+8)\ge 0$$
$$y^2 -10y + 25 -4y^2 +20y -32 \ge 0$$
which reduces to $$(y-1)\left(\frac{7}{3}-y\right)\ge 0 \implies 1 \le y\le 7/3$$
The solution to the orginal equation is
$$x = \frac{5-y \pm \sqrt{10y - 3y^2-7}}{2}$$
and so $$x^2 = \frac{(5-y)^2 + (10y - 3y^2-7) + 2(5-y)\sqrt{10y - 3y^2-7}}{4}$$
$$x^2 +y^2 = \frac{4y^2+(5-y)^2 + (10y - 3y^2-7) + 2(5-y)\sqrt{10y - 3y^2-7}}{4}$$
where $1\le y\le7/3$
some brute force differentiation shows that the function on the right achieves its maxima 8 at $y=2$
On
Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.
Hence, $v^2=4u^2-10u+8$.
But $4u^2-10u+8=v^2\leq u^2$, which gives $\frac{4}{3}\leq u\leq2$, and we need to prove that $$4u^2-2v^2\leq8$$ or $$2u^2-(4u^2-10u+8)\leq4$$ or $$u^2-5u+6\geq0$$ or $$(u-2)(u-3)\geq0,$$ which is true for $\frac{4}{3}\leq u\leq2$.
Done!
On
A simple solution with Vieta's relations:
Set $s=x+y,\;p=xy$. The relation between $x$ and $y$ rewrites as $p=s^2-5s+8$, so $$x^2+y^2=s^2-2p=-s^2+10s-16.$$ This is a quadratic polynomial in $s$. Its maximum is obtained for $s=5$, and it is equal to $9$.
Furthermore, the equation $\;t^2-st+p=0$ must have real roots, which means its discriminant $$\Delta(s)=s^2-4p=-3s^2+20s-32=(s-4)(8-3s)$$ is positive.
So the maximum of $f(s)=-s^2+10s-16$ on the domain for which $x$ and $y$ exist, is equal to $$\max(f(4),f(8/3))=f(4)=8.$$
Let $p=x+y$, $q=x-y$, then $$\frac{p+q}{2}\frac{p-q}{2}=p^2-5p+8$$ $$3p^2-20p+q^2+32=0$$ $$q^2=-3p^2+20p-32 \ge 0$$ So $$4\ge p \ge \frac{8}{3}$$ $$p^2+q^2=-2p^2+20p-32=18-2(p-5)^2 \le 18-2(4-5)^2=16$$ $$x^2+y^2=\frac{1}{2}(p^2+q^2)\le 8$$