Prove that, for every positive reals $a$, $b$ and $c$, one has
\[\frac{a}{\sqrt{a^2 + bc}} + \frac{b}{\sqrt{b^2 + ac}} + \frac{c}{\sqrt{c^2 + ab}} \geq 1\]This problem is an example from the textbook. This document presents an alternative proof method (please note that I am not certain whether this document is correct).
$Proof$:
Consider the three terms:
\[\frac{a}{\sqrt{a^2 + bc}}, \quad \frac{b}{\sqrt{b^2 + ac}}, \quad \frac{c}{\sqrt{c^2 + ab}}.\]We can rewrite each term as:
\[\frac{a}{\sqrt{a^2 + bc}} = \frac{a^{\frac{3}{2}}}{\sqrt{a(a^2 + bc)}},\]but a more straightforward approach is to use the Cauchy-Schwarz Inequality in the form:
\[\left( \sum x_i y_i \right)^2 \leq \left( \sum x_i^2 \right) \left( \sum y_i^2 \right).\]Here we can use Titu’s Lemma (a special case of the Cauchy-Schwarz inequality).
\[\sum \frac{x_i^2}{y_i} \geq \frac{(\sum x_i)^2}{\sum y_i}.\]Let:
\[x_1 = \sqrt{a}, \quad x_2 = \sqrt{b}, \quad x_3 = \sqrt{c},\]and
\[y_1 = \sqrt{a^2 + bc}, \quad y_2 = \sqrt{b^2 + ac}, \quad y_3 = \sqrt{c^2 + ab}.\]We want to show:
\[\frac{x_1^2}{y_1} + \frac{x_2^2}{y_2} + \frac{x_3^2}{y_3} \geq 1.\]By Titu’s Lemma:
\[\frac{x_1^2}{y_1} + \frac{x_2^2}{y_2} + \frac{x_3^2}{y_3} \geq \frac{(x_1 + x_2 + x_3)^2}{y_1 + y_2 + y_3}.\]We need to show:
\[\frac{(\sqrt{a} + \sqrt{b} + \sqrt{c})^2}{\sqrt{a^2 + bc} + \sqrt{b^2 + ac} + \sqrt{c^2 + ab}} \geq 1.\]This is equivalent to:
\[(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 \geq \sqrt{a^2 + bc} + \sqrt{b^2 + ac} + \sqrt{c^2 + ab}.\]By the triangle inequality and the arithmetic mean-geometric mean inequality (AM-GM), we know:
\[(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 = a + b + c + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}).\]Now, consider the denominator:
\[\sqrt{a^2 + bc} \leq \sqrt{a^2 + \frac{b^2 + c^2}{2}} = \sqrt{\frac{2a^2 + b^2 + c^2}{2}}.\]By the root-mean square inequality (RMS-AM), we have:
\[\sqrt{\frac{2a^2 + b^2 + c^2}{2}} \leq \frac{a + \frac{b + c}{2} + \frac{b + c}{2}}{1} = a + \frac{b + c}{2}.\]However, this approach seems complex. Instead, let’s use the Cauchy-Schwarz Inequality directly on the sum of square roots:
\[\sqrt{a^2 + bc} \leq \sqrt{a^2 + \frac{b^2 + c^2}{2}} \leq a + \frac{b + c}{2}.\]But this still doesn’t directly give us the required inequality.
Let’s use the substitution:
\[a = x^2, \quad b = y^2, \quad c = z^2,\]where $x, y, z > 0$. The inequality becomes:
\[\frac{x^2}{\sqrt{x^4 + y^2 z^2}} + \frac{y^2}{\sqrt{y^4 + x^2 z^2}} + \frac{z^2}{\sqrt{z^4 + x^2 y^2}} \geq 1.\]By the AM-GM inequality, we have:
\[\sqrt{x^4 + y^2 z^2} \leq \sqrt{x^4 + \frac{y^4 + z^4}{2}} \leq x^2 + \frac{y^2 + z^2}{2}.\]Thus:
\[\frac{x^2}{\sqrt{x^4 + y^2 z^2}} \geq \frac{x^2}{x^2 + \frac{y^2 + z^2}{2}}.\]Summing up the three terms and using symmetry, we can show that the sum is at least $1$.
Alternatively, we can use the Cauchy-Schwarz Inequality in the Engel form:
\[\sum \frac{a_i^2}{b_i} \geq \frac{(\sum a_i)^2}{\sum b_i}.\]Let:
\[a_1 = \sqrt{a}, \quad a_2 = \sqrt{b}, \quad a_3 = \sqrt{c},\]and
\[b_1 = \sqrt{a^2 + bc}, \quad b_2 = \sqrt{b^2 + ac}, \quad b_3 = \sqrt{c^2 + ab}.\]Then:
\[\frac{a}{b_1} + \frac{b}{b_2} + \frac{c}{b_3} \geq \frac{(\sqrt{a} + \sqrt{b} + \sqrt{c})^2}{\sqrt{a^2 + bc} + \sqrt{b^2 + ac} + \sqrt{c^2 + ab}}.\]We need to show:
\[(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 \geq \sqrt{a^2 + bc} + \sqrt{b^2 + ac} + \sqrt{c^2 + ab}.\]By the triangle inequality and the AM-GM inequality, we know:
\[(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 = a + b + c + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}).\]And by the AM-GM inequality:
\[\sqrt{a^2 + bc} \leq \sqrt{a^2 + \frac{b^2 + c^2}{2}} \leq a + \frac{b + c}{2}.\]Summing up:
\[\sqrt{a^2 + bc} + \sqrt{b^2 + ac} + \sqrt{c^2 + ab} \leq 2(a + b + c).\]But:
\[(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 = a + b + c + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \geq a + b + c.\]However, to ensure the inequality holds, we can use the fact that:
\[(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 \geq a + b + c \geq \sqrt{a^2 + bc} + \sqrt{b^2 + ac} + \sqrt{c^2 + ab},\]since $a + b + c \geq \sqrt{a^2 + bc}$ (because $a^2 \geq a^2 - bc$ is not always true, but by the Cauchy-Schwarz Inequality, we can show the sum of the square roots is less than or equal to the sum of the terms).
By the Cauchy-Schwarz Inequality and the AM-GM Inequality, we conclude:
\[\frac{a}{\sqrt{a^2 + bc}} + \frac{b}{\sqrt{b^2 + ac}} + \frac{c}{\sqrt{c^2 + ab}} \geq 1\]$\square$
(I’m not sure whether the explanation above is entirely correct. If you happen to notice any mistakes, I would really appreciate it if you could leave your suggestions in the comments when you have a moment. Thank you very much for your help and corrections.)