# Three Variables-Inequality with $a+b+c=abc$

$a$,$b$,$c$ are positive numbers such that $~a+b+c=abc$
Find the maximum value of $~\dfrac{1}{\sqrt{1+a^{2}}}+\dfrac{1}{\sqrt{1+b^{2}}}+\dfrac{1}{\sqrt{1+c^{2}}}$

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Trigonometric substitution looks good for this, especially if you know sum of cosines of angles in a triangle are $\le \frac32$. However if you want an alternate way…

Let $a = \frac1x, b = \frac1y, c = \frac1z$. Then we need to find the maximum of
$$F = \sum \frac{x}{\sqrt{x^2+1}}$$
with the condition now as $xy + yz + zx = 1$. Using this condition to homogenise, we have

\begin{align} F &= \sum \frac{x}{\sqrt{x^2 + xy + yz + zx}} \\ &= \sum \frac{x}{\sqrt{(x+y)(x+z)}} \\ &= \sum \frac{x\sqrt{(x+y)(x+z)}}{(x+y)(y+z)} \\ &\le \sum \frac{x\left((x+y)+(x+z)\right)}{2\cdot(x+y)(y+z)} \quad \text{by AM-GM}\\ &= \frac12 \sum \left(\frac{x}{x+y}+\frac{x}{x+z} \right) \\ &= \frac12 \sum \left(\frac{x}{x+y}+\frac{y}{y+x} \right) \quad \text{rearranging terms across sums}\\ &= \frac32 \end{align}

HINT:

Put $a=\tan A$ etc. so that $A+B+C=n\pi$

Now $\frac1{\sqrt{1+a^2}}=\cos A$

Use this

Hint：Let $a=\tan\alpha, b=\tan\beta, c=\tan\gamma$, then $a+b+c=abc$ implies $\alpha+\beta+\gamma=\pi$. So the objective function becomes $$\cos\alpha+\cos\beta+\cos\gamma=\cos\alpha+\cos\beta-\cos(\alpha+\beta)$$

From the condition, you can express “c” as a function of “a” and “b” and replace its value in the objective function F. Now, since you look for an extremum without any constraint, compute dF/da and dF/db and you want them to be zero. This gives you two equations (dF/da =0 and dF/db=0) to solve for “a” and “b”. The directly obtained expressions for dF/da and dF/db are not very nice but they simplify a lot. For example, you can extract “b” from dF/db and obtain : b = a / (Sqrt[a^2 + 1]-1). Use now dF/da=0 to solve for “a”; this corresponds to a simple expression and the solutions are a = 0 , a = Sqrt[3], a = – Sqrt[3]. Compute now the corresponding values of “b” and “c” and the value of the objective function.Since your unknowns are strictly positive, forget a=0.