Calculate the maximum value of $\sum_{cyc}\frac{1}{\sqrt{a^2 + b^2}}$ where $a, b, c > 0$ and $abc = a + b + c + 2$.

Question

$a$, $b$ and $c$ are positives such that $abc = a + b + c + 2$. Caculate the maximum value of $$\large \frac{1}{\sqrt{a^2 + b^2}} + \frac{1}{\sqrt{b^2 + c^2}} + \frac{1}{\sqrt{c^2 + a^2}}$$

This problem is adapted from a recent competition. Here's what I got, and I am pretty proud of it. (Actually not.)

$$abc = a + b + c + 2$$

$$\iff abc + (ab + bc + ca) + (a + b + c) + 1 = ab + bc + ca + 2(a + b + c) + 3$$

$$\iff (a + 1)(b + 1)(c + 1) = (a + 1)(b + 1) + (b + 1)(c + 1) + (c + 1)(a + 1)$$

$$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} = 1$$

And I'm done.

Answer 1

Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y},$ where $x$, $y$ and $z$ are positives.

Thus, $c=\frac{x+y}{z}$ and by AM-GM twice we obtain: $$\sum_{cyc}\frac{1}{\sqrt{a^2+b^2}}=\sum_{cyc}\frac{xy}{\sqrt{y^2(y+z)^2+x^2(x+z)^2}}\leq$$ $$\leq\sum_{cyc}\frac{xy}{\sqrt{2xy(x+z)(y+z)}}=\frac{1}{\sqrt2}\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq$$ $$\leq\frac{1}{2\sqrt2}\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{3}{2\sqrt2}.$$ The equality occurs for $x=y=z=1,$ which says that we got a maximal value.

Answer 2

So, by Cauchy-Schwarz, we have $a^2+b^2 \geq (a+b)^2/2$, so we just have to prove that

$\displaystyle \sum_{cyc} \frac{1}{a+b} \leq 3/4$ (It's obvious that this maximum occurs when $a=b=c=2$, I will prove it.)

Letting, $a=2k,b=2l,c=2m$ we have to prove that $\displaystyle \sum_{cyc} \frac{1}{k+l} \leq 3/2$ subject to the condition $4klm=k+l+m+1$.

By expanding and letting $k+l+m=p, kl+lm+mk, klm=r$ we have to prove that $3pq-3r \geq 2p^2+2q$.

Substituting $r=(p+1)/4$, we have to prove that $q(3p-2) \geq \frac{8p^2+3p+3}{4}$.

Now, it is known that $q^2 \geq 3pr=\frac{3p^2-3p}{4}$ so it suffices to show that $\sqrt{3p^2+3p}\cdot (3p-2) \geq \frac{8p^2+3p+3}{2}$ which is equivalent to :

$(p-3)(44p^3+48p^2-9p+3)$.

By AM-GM, $p+1=4r \leq 4p^3/27$ so easily $p \geq 3$.

So, $p-3 \geq 0$ and $44p^3+48p^2-9p+3 >48p^2-9p=p(48p-9)>0$ and the problem is solved.