Maximum value of $2\sqrt 6ab+8bc$ when $a^2+b^2+c^2=1$.

Question

Let $a,b,c$ be non-negative real numbers with $a^2+b^2+c^2=1$. What is the maximum value of $$2\sqrt 6ab+8bc?$$

I solved this problem in the following way:

By AM-GM, $$2\sqrt 6ab\leq\sqrt{22}a^2+\frac{3\sqrt2}{\sqrt{11}}b^2\\ 8bc\leq \frac{8\sqrt2}{\sqrt{11}}b^2+\sqrt{22}c^2$$ Summing them gives $$2\sqrt 6ab+8bc\leq \sqrt{22}(a^2+b^2+c^2)=\sqrt{22}$$ Equality holds when $(a,b,c)=\left(\frac{\sqrt3}{\sqrt{22}},\frac {1}{\sqrt2},\frac{2}{\sqrt{11}}\right)$.

Finding the coefficients is quite difficult in this solution. So, I want to know what other methods can be used to solve these kinds of problems.

Answer 1

Let $k$ be a maximal value.

Thus, $k>0$ and the inequality $$2\sqrt6ab+8bc\leq k(a^2+b^2+c^2)$$ or $$kb^2-2(\sqrt6a+4c)b+k(a^2+c^2)\geq0$$ is true for any reals $a$, $b$ and $c$,which says $$(\sqrt6a+4c)^2-k^2(a^2+c^2)\leq0$$ or $$(k^2-6)a^2-8\sqrt6ac+(k^2-16)c^2\geq0,$$ for which we need $k^2>16$ and $$96-(k^2-6)(k^2-16)=0,$$ which gives $$k^2(k^2-22)=0$$ or $$k=\sqrt{22}.$$

The equality occurs for $b=\frac{\sqrt6a+4c}{\sqrt22}$,$a=\frac{4\sqrt6b}{22-6}$ and $a^2+b^2+c^2=1,$ id est, occurs, which says that we got a maximal value.

Answer 2

Note that the objective function is $2b\cdot(\sqrt6a+4c)$, and we have $(\sqrt6a+4c)^2+(4a-\sqrt6c)^2=(\sqrt6^2+4^2)(a^2+c^2)=22(a^2+c^2)$, so \begin{align*} 2b\cdot(\sqrt6a+4c) &\leq\sqrt{22}\cdot\left[b^2+\left(\frac{\sqrt6a+4c}{\sqrt{22}}\right)^2\right]\\ &\leq\sqrt{22}\cdot\left[b^2+\left(\frac{\sqrt6a+4c}{\sqrt{22}}\right)^2+\left(\frac{4a-\sqrt6c}{\sqrt{22}}\right)^2\right] =\sqrt{22}(a^2+b^2+c^2)=\sqrt{22}. \end{align*} equality iff $4a-\sqrt6c=0$, $\frac{\sqrt6a+4c}{\sqrt{22}}=b=\frac1{\sqrt2}$.

Answer 3

Using Lagrange multipliers, you want to minimize $$F=2 \sqrt{6} a b+8 b c+\lambda \left(a^2+b^2+c^2-1\right)$$ So $$F'_a=2 a \lambda +2 \sqrt{6} b=0 \tag 1$$ $$F'_b=2 \sqrt{6} a+2 b \lambda +8 c=0 \tag 2$$ $$F'_c=8 b+2 c \lambda=0\tag 3$$ $$F'_\lambda=a^2+b^2+c^2-1=0\tag 4$$

From $(1)$, solve for $b$; from $(2)$, solve for $c$. Plug in $(3)$ to have $$\frac{a \lambda \left(\lambda ^2-22\right)}{2 \sqrt{6}}=0$$ So, $\lambda=\pm \sqrt{22}$. But the expressions are $$b=-\frac{a \lambda }{\sqrt{6}} \qquad \qquad c=\frac{a \left(\lambda ^2-6\right)}{4 \sqrt{6}}$$ Plug in $(4)$ $$\frac{1}{96} a^2 \left(\lambda ^4+4 \lambda ^2+132\right)=1$$ To have positive numbers, $\lambda=- \sqrt{22}$ which gives $$a=\sqrt{\frac{3}{22}}\qquad \qquad b=\frac{1}{\sqrt{2}}\qquad \qquad c=\frac{2}{\sqrt{11}}$$

Answer 4

Instead of everything AMGM, apply CS to $(2\sqrt{6},8)$ and $(a,c)$ to get $$|2\sqrt{6}a + 8c| \le \sqrt{(2\sqrt{6})^2 + 8^2}\sqrt{a^2+c^2} = \sqrt{88} \sqrt{a^2+c^2}$$

and AMGM to $(a^2+c^2,b^2)$ to get

$$\sqrt{a^2+c^2}|b| \le \frac12(a^2+c^2+b^2) = \frac12$$

Combine these give us

$$2\sqrt{6}ab + 8bc \le \sqrt{22}$$

To verify this is indeed the maximum, we know AMGM achieve equality when all items is the same. This means the $(a,b,c)$ which achieve maximum need to satisfy: $$a^2 + c^2 = b^2 = \frac12$$ WOLOG, let's take $b = \frac1{\sqrt{2}}$.

In order for CS to achieve equality, we need $(2\sqrt{6},8)$ and $(a,c)$ proportional to each other:

$$2\sqrt{6} : 8 = a : c$$

When $b > 0$, we need $a, c > 0$ to get a maximum. This suggests

$$\begin{align} a &= \frac{2\sqrt{6}}{\sqrt{88}}\sqrt{a^2+c^2} = \frac{2\sqrt{6}}{\sqrt{88}}\frac1{\sqrt{2}} = \sqrt{\frac{3}{22}}\\ c &= \frac{8}{\sqrt{88}}\sqrt{a^2+c^2} = \frac{8}{\sqrt{88}}\frac1{\sqrt{2}} = \frac{2}{\sqrt{11}} \end{align} $$

Substitute $(a,b,c)$ by $\left(\sqrt{\frac{3}{22}},\frac{1}{\sqrt{2}},\frac{2}{\sqrt{11}}\right)$ into $2\sqrt{6}ab + 8bc$ do give us the expected $\sqrt{22}$. So the maximum is indeed $\sqrt{22}$.

As one can see, figuring out a configuration $(a,b,c)$ which maximizes the expression isn't that difficult. It comes down to an appropriate decomposition of the problem into sub-problems. If you know the optimal configurations for the sub-problems, you can combine them to get an optimal configuration for original problem.

Answer 5

Well, here is a way to address finding such coeffts. Consider the following AM-GMs, with as yet undetermined $\alpha \in [0, 1]$.

$$a^2 + \alpha b^2 \geqslant 2\sqrt{\alpha}\,ab \tag1$$ $$(1-\alpha)b^2+c^2 \geqslant 2\sqrt{1-\alpha}\,bc \tag2$$

Adding the above, we get $$1=a^2+b^2+c^2 \geqslant 2\left(\sqrt{\alpha} \;ab + \sqrt{1-\alpha}\;bc \right)\tag{$\star$}$$

Comparing coefficients of the RHS with the expression we are trying to maximise, we need $\sqrt{\alpha} : \sqrt{1-\alpha} = \sqrt6 : 4 \iff \alpha:1-\alpha = 3:8 \iff \alpha = \frac3{11}$.

Using this value in $\star$ we get: $$\frac12 \geqslant \sqrt{\tfrac3{11}} \; ab + \sqrt{\tfrac8{11}} \; bc \implies \sqrt{22} \geqslant 2\sqrt6 \;ab + 8 bc$$

Equality is obviously possible for the two AM-GMs when $a^2:b^2:c^2=\alpha : 1 : 1-\alpha$, so this gives the maximum.

Answer 6

Put $a=\cos x\cos y, b=\sin y, c=\sin x\cos y$ You then get $$\sin y\cos y(2\sqrt 6\cos x+8\sin x)=\sin(2y)(\sqrt 6\cos x+4\sin x)$$ now the maximum is achived when both $\sin(2y)$ and $(\sqrt 6\cos x+4\sin x)$ are maximized.

This occurs when $\sin(2y)=1$ and $\cos x=\sqrt 6/4\sin x$ or $x=\arctan(4/\sqrt 6)$ coupled with $\sin(\arctan(x)) =\frac{x}{\sqrt{x^2+1}}$ as a disclaimer you'd need to assure the max happens for $a, b, c>0$ so $x, y\in(0,\pi/2)$, that the extrema is indeed a max not a min etc.