Does Trigonometric Substitution Works?

Relevant wiki: Hölder's Inequality

Rewrite the expression $\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}+\sqrt{\frac{1}{c^2}+\frac{2}{a^2}}+\sqrt{\frac{1}{b^2}+\frac{2}{c^2}}$ From the condition, we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ . So applying Minkowski Inequality $\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}+\sqrt{\frac{1}{c^2}+\frac{2}{a^2}}+\sqrt{\frac{1}{b^2}+\frac{2}{c^2}}\geq\sqrt{3\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)^2}=\sqrt{3}$ So $m=3$ and the equality holds when $\begin{cases} \frac{b^2}{2a^2}=\frac{a^2}{2c^2}=\frac{c^2}{2b^2} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\end{cases}$ or $a=b=c=3$

Obviously, $ab+bc+ca\quad =\quad abc\quad >\quad 0$ We have:

$S\quad =\quad \frac { \sqrt { b^{ 2 }+2{ a }^{ 2 } } }{ ab } +\frac { \sqrt { { a }^{ 2 }+2c^{ 2 } } }{ ac } +\frac { \sqrt { { c }^{ 2 }+2{ b }^{ 2 } } }{ bc } \\ \\ =\quad \frac { c\sqrt { b^{ 2 }+2{ a }^{ 2 } } +b\sqrt { { a }^{ 2 }+2c^{ 2 } } +a\sqrt { { c }^{ 2 }+2{ b }^{ 2 } } }{ abc } \\ \\ =\quad \frac { \sqrt { { b }^{ 2 }{ c }^{ 2 }+2{ a }^{ 2 }{ c }^{ 2 } } +\sqrt { { a }^{ 2 }{ b }^{ 2 }+2{ c }^{ 2 }{ b }^{ 2 } } +\sqrt { { c }^{ 2 }{ a }^{ 2 }+2{ b }^{ 2 }{ a }^{ 2 } } }{ abc }$ .

Applying the AM-GM inequality yields:

${ 2(b }^{ 2 }{ c^{ 2 }+{ a }^{ 2 }{ c }^{ 2 })\quad \ge \quad 4ab{ c }^{ 2 } }\\ \\ { a }^{ 2 }{ c }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 }\quad =\quad 2{ a }^{ 2 }{ c }^{ 2 }\\ \\ \Rightarrow \quad 2({ b }^{ 2 }{ c }^{ 2 }+2{ a }^{ 2 }{ c }^{ 2 })\quad \ge \quad 2{ a }^{ 2 }{ c }^{ 2 }+4ab{ c }^{ 2 }\\ \Rightarrow \quad 3({ b }^{ 2 }{ c }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 })\quad \ge \quad { b }^{ 2 }{ c }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 }+{ a }^{ 2 }{ c }^{ 2 }+2{ a }^{ 2 }{ c }^{ 2 }+2ab{ c }^{ 2 }+2{ a }^{ 2 }{ c }^{ 2 }+2ab{ c }^{ 2 }\quad =\quad (bc+2ac)^{ 2 }\\ \Rightarrow \quad { b }^{ 2 }{ c }^{ 2 }+{ 2a }^{ 2 }{ c }^{ 2 }\quad \ge \quad \frac { 1 }{ 3 } { (bc+2ac) }^{ 2 }\\ \Rightarrow \quad \sqrt { { b }^{ 2 }{ c }^{ 2 }+{ 2a }^{ 2 }{ c }^{ 2 } } \ge \quad \frac { \sqrt { 3 } }{ 3 } (bc+2ac)\quad \quad \quad \quad (because\quad bc+2ac\quad >\quad 0)\\ \\ Similarly,\quad \sqrt { { a }^{ 2 }{ b }^{ 2 }+2{ c }^{ 2 }{ b }^{ 2 } } \ge \quad \frac { \sqrt { 3 } }{ 3 } (ab+2bc)\\ \quad \qquad \qquad \quad \quad \sqrt { { c }^{ 2 }{ a }^{ 2 }+2{ b }^{ 2 }{ a }^{ 2 } } \ge \quad \frac { \sqrt { 3 } }{ 3 } (2ab+ac)\\ \\ Add\quad the\quad three\quad inequalities\quad side\quad to\quad side\quad yields\\ \\ \sqrt { { b }^{ 2 }{ c }^{ 2 }+{ 2a }^{ 2 }{ c }^{ 2 } } +\sqrt { { a }^{ 2 }{ b }^{ 2 }+2{ c }^{ 2 }{ b }^{ 2 } } +\sqrt { { c }^{ 2 }{ a }^{ 2 }+2{ b }^{ 2 }{ a }^{ 2 } } \ge \quad \frac { \sqrt { 3 } }{ 3 } 3(ab+bc+ac)\quad =\quad \sqrt { 3 } (ab+bc+ac)\quad =\quad \sqrt { 3 } abc\\ \Rightarrow \quad S\quad \ge \quad \frac { \sqrt { 3 } abc }{ abc } \quad =\quad \sqrt { 3 } \quad (abc>0)\\ \\ Equality\quad happens\quad when\quad a=b=c=3\\ So\quad m\quad =\quad 3$

Rewriting the expression into: $\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}+\sqrt{\frac{1}{c^2}+\frac{2}{a^2}}+\sqrt{\frac{1}{b^2}+\frac{2}{c^2}}$ OR $\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{b^2}}+\sqrt{\frac{1}{c^2}+\frac{1}{a^2}+\frac{1}{a^2}}+\sqrt{\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}}$ Rewriting the condition to $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} =1$ . According to Cauchy-Schwarz inequality, we may find that $\bigg(\frac{1}{a} + \frac{1}{b} +\frac{1}{b}\bigg)^2 \leq 3\bigg(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{b^2}\bigg)$ $\bigg(\frac{1}{c} + \frac{1}{a} +\frac{1}{a}\bigg)^2 \leq 3\bigg(\frac{1}{c^2} + \frac{1}{a^2} + \frac{1}{a^2}\bigg)$ $\bigg(\frac{1}{b} + \frac{1}{c} +\frac{1}{c}\bigg)^2 \leq 3\bigg(\frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{c^2}\bigg)$ OR $\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{b^2}} \geq \frac{1}{\sqrt{3}} \bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\bigg)$ $\sqrt{\frac{1}{c^2} + \frac{1}{a^2} + \frac{1}{a^2}} \geq \frac{1}{\sqrt{3}} \bigg(\frac{1}{c}+\frac{1}{a}+\frac{1}{a}\bigg)$ $\sqrt{\frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{c^2}} \geq \frac{1}{\sqrt{3}} \bigg(\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\bigg)$

Applying these to the expression, we get: $\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{b^2}}+\sqrt{\frac{1}{c^2}+\frac{1}{a^2}+\frac{1}{a^2}}+\sqrt{\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}} \geq \frac{1}{\sqrt{3}} \bigg(\frac{3}{a} +\frac{3}{b} + \frac{3}{c} \bigg) = \sqrt{3} \bigg(\frac{1}{a} + \frac{1}{b} +\frac{1}{c} \bigg) = \sqrt{3}$ which $m=3$

why to do such a large claculation of course that is a formal approach but there is a short trick to as it is mentions a,b,c are the positive real no just try to satisfy the relation $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ =1 you will find a=b=c=3

but you do learn the formal because it is more important than since it opens your mind to possiblitys