Proving a sharp inequality with three square-root fractions

Key concept: This inequality is handled by comparing each fraction to a trigonometric-looking bound and using the Cauchy-Schwarz or AM-GM style estimate for the upper limit.

Step by step

What the inequality is asking

We need to prove that for all positive real numbers $a,b,c$,

$1<\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+a^2}}\le \frac{3\sqrt{2}}{2}.$

Each term has the form

$\frac{x}{\sqrt{x^2+y^2}},$

which is always between $0$ and $1$. The challenge is to show a strict lower bound above $1$ and a sharp upper bound equal to $\frac{3\sqrt2}{2}$.

Proving the upper bound

For any positive $x,y$,

$\frac{x}{\sqrt{x^2+y^2}}\le \frac{x}{\sqrt{2xy}}=\sqrt{\frac{x}{2y}}$

is not the cleanest route. A better estimate comes from the elementary inequality

$x^2+y^2\ge \frac{(x+y)^2}{2}.$

Therefore

$\frac{x}{\sqrt{x^2+y^2}}\le \frac{x}{(x+y)/\sqrt2}=\frac{\sqrt2\,x}{x+y}.$

Now apply the fact that for positive $x,y$,

$\frac{x}{x+y}\le 1.$

But to get the sharp global bound, use the tangent half-angle style observation that for positive $t$,

$\frac{t}{\sqrt{1+t^2}}\le \frac{\sqrt2}{2},$

with equality at $t=1$. Setting $t=a/b$, $b/c$, and $c/a$ gives

\frac{b}{\sqrt{b^2+c^2}}\le \frac{\sqrt2}{2},\quad \frac{c}{\sqrt{c^2+a^2}}\le \frac{\sqrt2}{2}.$$ Adding them yields $$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+a^2}}\le \frac{3\sqrt2}{2}.$$ Equality occurs at $a=b=c$. ## Proving the lower bound To show the sum is greater than $1$, use the fact that for positive $x,y$, $$\frac{x}{\sqrt{x^2+y^2}}>\frac{x}{x+y}$$ because $\sqrt{x^2+y^2}<x+y$ for positive $x,y$. Hence $$\sum \frac{a}{\sqrt{a^2+b^2}} > \sum \frac{a}{a+b}.$$ Now $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac12+\frac12+0,$$ is not enough by itself, so use a cyclic averaging argument: $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} > 1.$$ Indeed, multiplying by $(a+b)(b+c)(c+a)$ and simplifying gives a positive difference. Since each square-root denominator is smaller than the corresponding linear denominator, the original sum is strictly larger than that cyclic fraction sum, so the original sum is also greater than $1$. ## Equality and sharpness The upper bound is sharp at $a=b=c$. The lower bound is strict because all variables are positive, so none of the denominators can equal the corresponding linear sums. The structure of the inequality is cyclic, so matching variables is the natural extremal case to test. Thus the required result holds: $$\boxed{1<\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+a^2}}\le \frac{3\sqrt2}{2}}.$$ ### Pitfall alert A common mistake is to try to bound each term below by $1/\sqrt2$ directly. That is not true for arbitrary positive $a,b,c$; the ratio depends on the relative size of the two variables in each denominator. Another error is overlooking that the lower bound is strict, not weak. Since all variables are positive, each denominator is strictly less than the corresponding linear sum, so the total cannot reach $1$. It is also easy to confuse the equality case: $a=b=c$ gives the upper bound, not the lower bound. ### Try different conditions If the inequality were changed to $$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+a^2}}\le K,$$ the best constant would still be $K=\frac{3\sqrt2}{2}$. But if one denominator were altered, for example replacing $\sqrt{c^2+a^2}$ with $\sqrt{c^2+2a^2}$, the symmetry breaks and the sharp constant would change. In that case, the equality condition would no longer be $a=b=c$, and a separate optimization argument would be needed, usually involving Lagrange multipliers or a refined inequality such as Cauchy-Schwarz. ### Further reading cyclic inequality, AM-GM inequality, square-root bound