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Let $x,y,z\in\mathbb{R}$.Let $xy+yz+xz=1$.

Prove:$\displaystyle \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\leq \frac{3}{2}$

Zhipu 'Wilson' Zhao
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2 Answers2

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Hint: $$x^2 + 1 = x^2 + xy + yz + zx = (x+y)(x+z)$$ Also, you will need the inequality $$ \sqrt{AB}\le \frac12(A+B) $$

Solution:

$$ \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}} \\= \frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(x+z)}} +\frac{z}{\sqrt{(x+y)(x+z)}} \\ \le \frac12\left[ \frac x{x+y} + \frac x{x+z} + \frac y{y+z} + \frac y{x+y} + \frac z{z+y} + \frac z{z+x} \right]=3/2 $$

mookid
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    How do we know that $\frac{x}{x+y}$, etc., are positive (so we can apply AM/GM)? Example: $x=-1,y=z=1+\sqrt2$. –  Oct 16 '14 at 23:55
  • I did not think about that. I guess we should start by proving that the argmax is positive. – mookid Oct 16 '14 at 23:58
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Let $a,b,c\in(0,\pi)$ be such that $x=\cot a$, $y=\cot b$, $z=\cot c$.

Using the addition formula for cotangent, one can show that $$ \cot(a+b+c) = \frac{\cot a\cot b\cot c - \cot a - \cot b - \cot c} {\cot a\cot b + \cot b\cot c + \cot c\cot a - 1} \tag{$\ast$} $$ By hypothesis, the denominator on the RHS is 0, so (see below) the LHS is $\infty$; since $a,b,c\in (0,\pi)$, this implies $a+b+c\in\{\pi,2\pi\}$.

Case $a+b+c=\pi$: Then at most one of $a,b,c$ is greater than $\frac\pi2$; wlog, $a,b\in(0,\frac\pi2]$. Since cosine is concave on that interval, we have \begin{align*} \frac{x}{\sqrt{x^2+1}} + \frac{y}{\sqrt{y^2+1}} + \frac{z}{\sqrt{z^2+1}} &= \cos a + \cos b + \cos c \\ &\le 2\cos(\tfrac{a+b}{2}) + \cos c \\ &= 2\cos(\tfrac{a+b}{2}) - \cos(a+b) \\ &= 2\cos(\tfrac{a+b}{2}) - 2\cos^2(\tfrac{a+b}{2}) + 1 \\ &= \tfrac32 - 2\big(\cos(\tfrac{a+b}{2}) - \tfrac12\big)^2 \\ &\le \tfrac32 \end{align*} with equality iff $a=b=\frac\pi3$ (whence $c=\frac\pi3$ also), that is, $x=y=z=\frac1{\sqrt3}$.

Case $a+b+c=2\pi$: Then at most one of $a,b,c$ is less than $\frac\pi2$; wlog, $a,b\in[\frac\pi2,\pi)$. Since cosine is nonpositive in that interval, we have $$ \frac{x}{\sqrt{x^2+1}} + \frac{y}{\sqrt{y^2+1}} + \frac{z}{\sqrt{z^2+1}} = \cos a + \cos b + \cos c \le \cos c \le 1 < \tfrac32 $$

Now, one annoying detail about ($\ast$): I noted that the denominator on the RHS is zero, and inferred that the RHS, and hence the LHS, is $\infty$. But what if the numerator on the RHS is also zero? Well, suppose for contradiction that both numerator and denominator are zero. Then $xyz=x+y+z$ and $xy+yz+zx=1$, and so \begin{align*} (t-x)(t-y)(t-z) &= t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz \\ &= t^3 - xyzt^2 + t - xyz \\ &= (t^2+1)(t-xyz) \end{align*} which is impossible because the first polynomial has three real roots (counting multiplicity) but the last has only one.