# Let $a,b\in \mathbb{R}$ such that $ab=2$. Find the max value of $\frac{3}{2\left(a+b\right)^2}$ and $a, b$ where max is attained, without calculus.

My thinking:

By Arithmetic and Geometric mean inequality (AGM):

$$ab\le \left(\frac{a+b}{2}\right)^2$$

We know $$ab=2$$

$$\rightarrow$$ $$2\le \left(\frac{a+b}{2}\right)^2$$

$$\rightarrow$$ $$2\le \frac{a^2+b^2}{4}$$

$$\rightarrow$$ $$\frac{1}{2}\le \:\frac{1}{\frac{a^2+b^2}{4}}$$

$$\rightarrow$$ $$\frac{1}{2}\le \:\frac{4}{a^2+b^2}$$

$$\rightarrow$$ $$\frac{1}{2}\left(\frac{3}{a^2+b^2}\right)\le \:\frac{4}{a^2+b^2}\left(\frac{3}{a^2+b^2}\right)$$

$$\rightarrow$$ $$\frac{3}{2\left(a+b\right)^2}\le \frac{12}{\left(a+b\right)^4}$$

Therefore the max value is $$\frac{12}{\left(a+b\right)^4}$$

Maximum value is attained when $$a=b$$:

So we can write:

$$\rightarrow$$ $$\frac{3}{2\left(a\right)^2}=\frac{12}{\left(a\right)^4}$$

$$\rightarrow$$ $$3a^4=24a^2$$

$$\rightarrow$$ $$a=0, -2\sqrt{2},+2\sqrt{2}$$

Therefore the max value of $$a,b$$ are $$0,-2\sqrt{2},+2\sqrt{2}$$

I'm not completely sure about my answer, if anyone could provide some feedback, that would be amazing! Thanks in advance.

EDIT:

$$ab\le \frac{\left(a+b\right)^2}{2}\:\rightarrow \:2\le \frac{\left(a+b\right)^2}{4}\:\rightarrow \:\frac{1}{2}\le \frac{\left(a+b\right)^2}{16}\:\rightarrow \:\frac{3}{2\left(a+b\right)^2}\le \:\frac{3}{16}$$

• $(a + b)^2 \ne a^2 + b^2$, $(a + b)^2 = a^2 + 2ab + b^2$. See en.wikipedia.org/wiki/Freshman%27s_dream.
– Gary
Jan 14, 2022 at 2:02
• Note that AM-GM requires you to assume $a,b \geq 0$. This is actually sufficient to find the maximum, but in a proof you should show why you may reduce to this case. Jan 14, 2022 at 2:05
• @Yaya123 Note your second line is $2\le \frac{a^2+b^2}{4}$ while your third is $\frac{1}{2}\le \:\frac{1}{\frac{a^2+b^2}{4}}$. However, with both positive (or negative) values, when taking reciprocals of both sides in an inequality, the inequality direction changes. For example, by dividing both sides by $ab$, we get $a \le b \; \to \; \frac{1}{b} \le \frac{1}{a} \; \to \; \frac{1}{a} \ge \frac{1}{b}$. Jan 14, 2022 at 6:46

Why you found the maximal value of $$a, b$$? The question is asking $$\frac3{2(a+b)^2}$$!

Using AM-GM is, yes, the best approach. Here is my answer.

Since using AM-GM, $$8=4ab\le(a+b)^2$$, so $$\frac1{(a+b)^2}\le\frac18$$.

So the maxima of $$\frac3{2(a+b)^2}$$ is $$\frac32\cdot\frac18=\frac3{16}$$ when $$a=b=\pm\sqrt2$$.

• Hey I found another way to do it without AGM, would this be valid: $ab\le \frac{\left(a+b\right)^2}{2}\:\rightarrow \:2\le \frac{\left(a+b\right)^2}{4}\:\rightarrow \:\frac{1}{2}\le \frac{\left(a+b\right)^2}{16}\:\rightarrow \:\frac{3}{2\left(a+b\right)^2}\le \:\frac{3}{16}$ Jan 14, 2022 at 19:20

Substitute the given restriction into the proposed expression.

Then your problem reduces to study a single-variable real-valued function: \begin{align*} \frac{3}{2(a + b)^{2}} & = \frac{3}{2\left(a + \dfrac{2}{a}\right)^{2}}\\\\ & = \frac{3a^{2}}{2(a^{2} + 2)^{2}}\\\\ & = \frac{3}{2}\left(\frac{a}{a^{2} + 2}\right)^{2} \end{align*}

Since the quadratic function is increasing, it suffices to study when the argument attains its max.

More precisely, according to the AM-GM inequality, we can conclude that \begin{align*} a^{2} + 2 \geq 2\sqrt{2a^{2}} = 2\sqrt{2}|a| \Longleftrightarrow \left|\frac{a}{a^{2} + 2}\right| \leq \frac{1}{2\sqrt{2}} \end{align*}

Hence the maximum value attained is given by: \begin{align*} \frac{3}{2(a + b)^{2}} \leq \frac{3}{2}\left(\frac{1}{2\sqrt{2}}\right)^{2} = \frac{3}{16} \end{align*}

In order to determine $$a$$ which corresponds to the maximum, you can solve the equation: \begin{align*} a^{2} - 2\sqrt{2}|a| + 2 = 0 \end{align*}

Once you know $$a$$, you also know $$b$$ and you are done.

Hopefully this helps !