Trigonomean

Using the tangent addition formula, $g(x, y) = \tan(\tan^{-1}(x) + \tan^{-1}(y))$ $=$ $\frac{\tan(\tan^{-1}(x)) + \tan(\tan^{-1}(y))}{1 - \tan(\tan^{-1}(x))\tan(\tan^{-1}(y))}$ $=$ $\frac{x + y}{1 - xy}$ .

Therefore, $f_{TM}(x, y) = -\frac{1+\sqrt{1+g^2(x,y)}}{g(x,y)}$ $=$ $-\frac{1+\sqrt{1+(\frac{x + y}{1 - xy})^2}}{\frac{x + y}{1 - xy}}$ which simplifies to $f_{TM}(x, y) = \frac{xy - 1 + \sqrt{x^2y^2 + x^2 + y^2 + 1}}{x + y}$ .

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Now $f_{QM} \geq f_{GM}$ , so for two positive real numbers $x$ and $y$ , $\sqrt{\frac{x^2 + y^2}{2}} \geq \sqrt{xy}$ . Then if $xy > 1$ ,

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

$\frac{x^2 + y^2}{2} \geq xy$

$x^2 + y^2 \geq 2xy$

$x^2y^2 + x^2 + y^2 + 1 \geq x^2y^2 + 2xy + 1$

$x^2y^2 + x^2 + y^2 + 1 \geq (xy + 1)^2$

$\sqrt{x^2y^2 + x^2 + y^2 + 1} \geq xy + 1$

$xy + \sqrt{x^2y^2 + x^2 + y^2 + 1} \geq 2xy + 1$

$xy - 1 + \sqrt{x^2y^2 + x^2 + y^2 + 1} \geq 2xy$

$\frac{xy - 1 + \sqrt{x^2y^2 + x^2 + y^2 + 1}}{x + y} \geq \frac{2xy}{x + y}$

$\frac{xy - 1 + \sqrt{x^2y^2 + x^2 + y^2 + 1}}{x + y} \geq \frac{2}{\frac{1}{x} + \frac{1}{y}}$

$f_{TM} \geq f_{HM}$

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Also, $f_{AM} \geq f_{GM}$ , so for two positive real numbers $x$ and $y$ , $\frac{x + y}{2} \geq \sqrt{xy}$ . Then if $xy > 1$ ,

$\frac{x + y}{2} \geq \sqrt{xy}$

$x + y \geq 2\sqrt{xy}$

$(x + y)(xy - 1) \geq 2\sqrt{xy}(xy - 1)$

$xy(x + y) - (x + y) \geq 2xy\sqrt{xy} - 2\sqrt{xy}$

$xy(x + y)(x + y) - (x + y)(x + y) \geq 2xy\sqrt{xy}(x + y) - 2\sqrt{xy}(x + y)$

$xy(x + y)^2 - (x + y)^2 \geq 2xy\sqrt{xy}(x + y) - 2\sqrt{xy}(x + y)$

$xy(x + y)^2 - 2xy\sqrt{xy}(x + y) + 2\sqrt{xy}(x + y) \geq (x + y)^2$

$xy(x^2 + 2xy + y^2 - 2x\sqrt{xy} - 2y\sqrt{xy}) + 2\sqrt{xy}(x + y) \geq x^2 + 2xy + y^2$

$xy(x^2 + 2xy + y^2 - 2x\sqrt{xy} - 2y\sqrt{xy}) + 2\sqrt{xy}(x + y) - 2xy \geq x^2 + y^2$

$xy(x^2 + 2xy + y^2 - 2x\sqrt{xy} - 2y\sqrt{xy}) + 2\sqrt{xy}(x + y) - 2\sqrt{xy}\sqrt{xy} \geq x^2 + y^2$

$xy(x^2 + 2xy + y^2 - 2x\sqrt{xy} - 2y\sqrt{xy}) + 2\sqrt{xy}(x + y - \sqrt{xy}) \geq x^2 + y^2$

$x^2y^2 + xy(x^2 + 2xy + y^2 - 2x\sqrt{xy} - 2y\sqrt{xy}) + 2\sqrt{xy}(x + y - \sqrt{xy}) + 1 \geq x^2y^2 + x^2 + y^2 + 1$

$xy(x^2 + 3xy + y^2 - 2x\sqrt{xy} - 2y\sqrt{xy}) + 2\sqrt{xy}(x + y - \sqrt{xy}) + 1 \geq x^2y^2 + x^2 + y^2 + 1$

$xy(x + y - \sqrt{xy})^2 + 2\sqrt{xy}(x + y - \sqrt{xy}) + 1 \geq x^2y^2 + x^2 + y^2 + 1$

$(\sqrt{xy}(x + y - \sqrt{xy}) + 1)^2 \geq x^2y^2 + x^2 + y^2 + 1$

$\sqrt{xy}(x + y - \sqrt{xy}) + 1 \geq \sqrt{x^2y^2 + x^2 + y^2 + 1}$

$\sqrt{xy}(x + y) - xy + 1 \geq \sqrt{x^2y^2 + x^2 + y^2 + 1}$

$\sqrt{xy}(x + y) \geq xy - 1 + \sqrt{x^2y^2 + x^2 + y^2 + 1}$

$\sqrt{xy} \geq \frac{xy - 1 + \sqrt{x^2y^2 + x^2 + y^2 + 1}}{x + y}$

$f_{GM} \geq f_{TM}$

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Since $f_{GM} \geq f_{TM}$ and $f_{TM} \geq f_{HM}$ , and $f_{QM} \geq f_{AM} \geq f_{GM} \geq f_{HM}$ , $\boxed{f_{QM} \geq f_{AM} \geq f_{GM} \geq f_{TM} \geq f_{HM}}$ .

Well actially if you are a genius problem solver you can solve this really quickly without solving it mathematically. You just go by pin-pointing by similarity, out of possible answers formulated as abcde abdce adcbe abdec adcbe you have most primary similarities three, then you cut of from there ABdec, ABcde, ABdce and you eliminate by separating secondary similarities Dec and Dce, then because there is only two and only a switch between terms you don't have anymore to separate so you use an algorithm by noticing in the previous answers which term is most interchanged and never repeats or mrepeats less than the other comparative term (E) on the same spot for instance if C never repeats on the same spot and always interchanges it will be Ce, but if both C and E repeat on same locations you just count which one repeats on less locations over the ratio of those groups entering primary, secondary phases.