For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$.

Question

For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$. My attempt: We have $$2(a^2+b^2)\geq (a+b)^2$$ so $$-2\leq a+b \leq 2$$ In other hand $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$

Answer 1

Observe, $$a^2+b^2=2\implies (a+b)^2=2(ab+1)$$ It remains to show that $$3(a+b)+(ab+1)+4\ge0\;\Longleftrightarrow\;3(a+b)+\frac{(a+b)^2}{2}+4\ge0$$ Let $a+b=x$, we have, $$3x+\frac{x^2}{2}+4\ge0\;\Longleftrightarrow\;x^2+6x+8\ge0\;\Longleftrightarrow\;(x+2)(x+4)\ge0$$ $$\therefore\;x\le-4 \;\;\text{or}\;\;x\ge -2$$ Since you have shown that $a+b\ge-2$, the inequality must be true.

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Alternate Solution: Using AM-GM inequality, $$\frac{a^2+b^2}{2}\ge\sqrt{a^2b^2}\implies 1\ge ab$$ Adding the inequalities, $$3a+3b+ab\ge-5$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\ge ab$$ It remains to show that $a+b\ge-2$ which we know is true.

Answer 2

This is almost certainly not the intended method (Which would involve algebra), but you can always break out calculus to solve this. Effectively we are looking to minimize the function $f(x,y)=3x+3y+xy$ on the circle $x^2+y^2=2$. Since the circle is closed and bounded, it is compact and therefore we are guaranteed extreme values. Using Lagrange multipliers, we know that introducing a variable $\lambda$, the maximum and minimum must occur at points that satisfy $$\nabla f=\lambda \nabla g$$ where $g$ is your constraint.

Thus, taking partials with respect to $a$ and $b$ and setting them equal to each other, we get $$3+b=2\lambda a$$$$3+a=2\lambda b$$ assuming $a\neq 0, b\neq 0$ and solving for $\lambda$ in each and setting them equal to each other, we get $$\frac {3+b} a=\frac {3+a} b$$ Cross multiplying, we get $$3b+b^2=3a+a^2$$

Since quadratics are 2 to one (except for x=0), we get the only solutions here are $a=\pm b$. Combine that with our constraint equation $a^2+b^2=2$ gets you possible critical values of $(\pm 1,\pm 1)$ for the non equal to 0 case. Just trying out the equal to 0 cases adds the points $(0,\pm \sqrt 2),(\pm \sqrt 2,0)$

Evaluating at all the cases, you get the smallest value occurs at $(-1,-1)$ which is $-5$, thus you have the inequality in question. (Plus you even know the bound is the achieved and where)

Answer 3

Let $x=a+1,y=b+1$. Then the statement is equivalent to $$(x-1)^2+(y-1)^2=2\Rightarrow 3(x-1)+3(y-1)+(x-1)(y-1)\geq -5,$$ or $$x^2+y^2=2x+2y\Rightarrow (2x+2y)+xy\geq 0,$$ which is clear, since $$(x^2+y^2)+xy\geq 0.$$

Answer 4

Let $a = \sqrt{2} \cos u, b = \sqrt{2} \sin u$ which immediately satisfies the condition. Using Simon's Favorite Factoring Trick, $9 + 3a + 3b + ab = (a + 3)(b + 3)$, so by C-S:

$$(\sqrt{2} \cos u + 3)(\sqrt{2} \sin u + 3) ≥(\sqrt{\sqrt{2} \cos u} \sqrt{\sqrt{2} \sin u} + \sqrt{3} \sqrt{3})^2$$

$$= (\sqrt{2 \cos u \sin u} + 3)^2 = (\sqrt{\sin(2u)} + 3)^2$$

There are two values which have $\sqrt{2} \cos u$, $\sqrt{2} \sin u$ as their square. Choosing the negative value for the minimum, this is greater or equal to $(-1 + 3)^2 = 4$, hence $3a + 3b + ab ≥ 4 - 9 = -5$.

Answer 5

let $a=x-y$ and $b=x+y$

$a^2+b^2=2 \implies x^2+y^2=1 $

substituting $y$ given that $x^2+y^2=1 \implies $ the problem now is to prove that $6x+x^2-y^2 \ge -5 \iff 2x^2+6x-1 \ge -5 $ (1)

$y^2=1-x^2 \ge 0 \implies -1 \ge x \ge 1$ it's enough to see that (1) $\iff (x-1)(2x-4) $ which is easy to check that it's positive when $-1 \ge x \ge 1$