Why I love inequality so much

With $x=\tan u$ , $y=2\tan v$ , $z=3\tan w$ , we need to maximize $\sin u \sqrt{\sin v\sin w}$ over $0 < u,v,w < \tfrac12\pi$ and $u+v+w=\pi$ . With $u+v+w=\pi$ we have $\begin{array}{rcl} \sin u \sqrt{\sin v \sin w} & = & \sin u \sqrt{\frac{\cos(v-w) -\cos(v+w)}{2}} \;=\; \sin u \sqrt{\frac{\cos(v-w) +\cos u}{2}} \\ &\le& \sin u \sqrt{\frac{1+\cos u}{2}} \;=\; \sin u \cos \tfrac12u \\ &=& 2\sin\tfrac12u \cos^2\tfrac12u \end{array}$ Simple calculus tells us that this last expression is maximized over $0<u<\tfrac12\pi$ when $\tan \tfrac12u =\tfrac{1}{\sqrt{2}}$ . Thus the maximum value is $\tfrac{4}{3\sqrt{3}}=0.7698$ . This value is achieved when $v=w=\tfrac12(\pi-u)$ and $u = 2\tan^{-1}\tfrac{1}{\sqrt{2}}$ .

Let $x = \tan(a), y = 2\tan(b)$ and $z = 3\tan(c)$ , the equation now becomes $\begin{aligned} \frac{x\sqrt{yz}}{\sqrt{x^2 + 1} \times \sqrt[4]{(y^2 + 4)(z^2 + 9)}} &= \frac{\tan a\sqrt{6\tan b \tan c}}{\sqrt{\tan^2 a + 1} \times \sqrt[4]{(4\tan^2 b + 4)(9\tan^2 c + 9)}} \\ &= \frac{\tan a\sqrt{\tan b \tan c}}{\sec a \times \sqrt{\sec b \sec c}} \\ &= \sin{a} \sqrt{\sin b \sin c} \end{aligned}$

In addition to this, we also have the requirement that $6x + 3y + 2z = xyz$ or in this case $\begin{aligned} 6x + 3y + 2z = xyz & \implies 6\tan a + 6\tan b + 6 \tan c = 6\tan a \tan b \tan c \\ & \implies \tan b + \tan c = \tan a (\tan b \tan c - 1) \\ & \implies - \frac{\tan b + \tan c}{1 - \tan b \tan c} = \tan a \\ & \implies - \tan(b+c) = \tan a \\ & \implies a = \pi - (b+c) \\ & \implies a + b + c = \pi \end{aligned}$

Now we just need to find the maximum of $\sin(b+c) \sqrt{\sin b \sin c}$ . Mark's solution does this so you should read his.

X=y= 281/100. z=424/ 100 plugging in these values you have ur answer

Extremum with constraints will be found via Lagrange multipliers. Using a program solving non-linear equations' systems I got:

x = 2,828427124746

y = 2,828427124746

z = 4,242640687119

With these values I determined the maximum: $0.769800$

$\boxed{Remark:}$ To show that we have a maximum at position (x, y, z) refer to

$http://www.math.northwestern.edu/~clark/285/2006-07/handouts/lagrange-2deriv.pdf$

which presents the algorithm via Hessian bordered marix additionally by means of examples.