An Inequality with Just Two Variables III

In a right triangle with sides $a,b,\sqrt{a^2+b^2},\$ let $\theta be the acute angle opposite $a:\,$ $a=\sqrt{a^2+b^2}\sin\theta,\,$ $a:\,$ $b=\sqrt{a^2+b^2}\cos\theta.\,$ Then

$\displaystyle LHS=\frac{1}{\sqrt{2}}\tan\theta+\sqrt{2}\cot\theta+2\cos\theta+2\sec\theta\overset{def}{=}f(\theta).$

Solving

$\displaystyle f'(\theta)=\frac{1}{\sqrt{2}\cos^2\theta}-\sqrt{2}\frac{1}{\sin^2\theta}-2\sin\theta+\frac{2\sin\theta}{cos^2\theta}=0.$

we get successively

$\displaystyle\frac{\sin^2\theta-2\,\cos^2\theta}{\sqrt{2}\cos^2\theta\,\sin^2\theta}-2\sin\theta\frac{\cos^2\theta-1}{\cos\theta}=0,$

$\displaystyle\frac{3\,\sin^2\theta-2}{\sqrt{2}\,\cos^2\theta\,\sin^2\theta}+2\frac{\sin^3\theta}{\cos\theta}=0,$

$\displaystyle \frac{2-3\,\sin^2\theta}{\sqrt{2}\,\cos^2\theta\,\sin^2\theta}=2\frac{\sin^3\theta}{\cos\theta},$

$\displaystyle 2-3\,\sin^2\theta=2\sqrt{2}\,\sin^5\theta,$

$\displaystyle 4+9\,\sin^4\theta -12\,\sin^2\theta=8\,\sin^10\theta.$

With $t=\sin^2\theta\gt 0,$

$\displaystyle 8t^5-9t^4+12t-4=0,$

$\displaystyle (2t-1)(4t^4+2t^3+(t-2)^2=0,$

$t=\displaystyle\frac{1}{2},\,$ $\sin\theta=\displaystyle\frac{\sqrt{2}}{2},\,$ $\displaystyle\theta=\frac{\pi}{4}.$

Now,

$\displaystyle\begin{align} f''(\theta)=\sqrt{2}\,\sec^2\theta\,\tan\theta+2\sqrt{2}\,\frac{\cot\theta}{\sin^2\theta}-2\,\cos\theta+2\,\sec^3\theta+2\,\tan^2\theta\,\sec\theta. \end{align}$

$\displaystyle f''\left(\frac{\pi}{4}\right)=11\sqrt{2}>0,\,$ implying that $\displaystyle\theta=\frac{\pi}{4}$ is a minimum and that $f\,$ never attends a maximum on $\displaystyle\left(0,\frac{\pi}{2}\right).\,$ Hence,

$\displaystyle f(\theta)\ge f\left(\frac{\pi}{2}\right)=4\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{9\sqrt{2}}{2}.$

Note that $a\sin x+b\cos x\le\sqrt{a^2+b^2}\,$ so that

$\displaystyle\frac{a\sin x}{\sqrt{a^2+b^2}}+\frac{b\cos x}{\sqrt{a^2+b^2}}\le 1,$

with equality only when $a=\sqrt{a^2+b^2}\sin x,\,$ $b=\sqrt{a^2+b^2}\cos x,\,$ $x\in\displaystyle\left(0,\frac{pi}{2}\right).\,$ The given inequality is equivalent to

$\displaystyle f(x)=\frac{1}{\sqrt{2}}\tan x+\sqrt{2}\cot x +\frac{2}{\cos x}+2\,\cos x\ge\frac{9\sqrt{2}}{2}.$

Where's $f(x)=0?

$\displaystyle f'(x)=\frac{1}{\sqrt{2}\,\cos^2 x}+\frac{\sqrt{2}}{\sin^2 x} -\frac{2\,\sin x}{\cos^2 x}-2\,\sin x=0,\,$

or, $(2\sqrt{2}\sin^3x-1)(\cos^2x+1).\,$ $(2\sqrt{2}\sin^3x-1)=0\,$ implies $x=\displaystyle\frac{\pi}{4}.\,$ $\displaystyle f\left(\frac{\pi}{4}\right)=\frac{9\sqrt{2}}{2}.$

Using the AM-GM inequality,

$\displaystyle\begin{align} 3&\cdot\frac{\displaystyle\frac{a}{b\sqrt{2}}+\frac{b}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{a^2+b^2}}}{3}+3\cdot\frac{\displaystyle\frac{b\sqrt{2}}{a}+\frac{\sqrt{a^2+b^2}}{b}+\frac{\sqrt{a^2+b^2}}{b}}{3}\\ &\qquad=3\left(\frac{ab}{\sqrt{2}(a^2+b^2)}\right)^{\frac{1}{3}}+6\left(\frac{\sqrt{2}(a^2+b^2)}{8ab}\right)^{\frac{1}{3}}\\ &\qquad=3\cdot 3\cdot\frac{\displaystyle\left(\frac{ab}{\sqrt{2}(a^2+b^2)}\right)^{\frac{1}{3}}+\left(\frac{\sqrt{2}(a^2+b^2)}{8ab}\right)^{\frac{1}{3}}+\left(\frac{\sqrt{2}(a^2+b^2)}{8ab}\right)^{\frac{1}{3}}}{3}\\ &\qquad\ge 9\cdot\left[\left(\frac{ab}{\sqrt{2}(a^2+b^2)}\right)^{\frac{1}{3}}\left(\frac{\sqrt{2}(a^2+b^2)}{8ab}\right)^{\frac{2}{3}}\right]^{\frac{1}{3}}\\ &\qquad =9\cdot\left[\left(\frac{a^2+b^2}{ab}\right)^{\frac{1}{3}}\cdot\left(\sqrt{2}\right)^{\frac{2}{3}-\frac{1}{3}-4}\right]^{\frac{1}{3}}\\ &\qquad\ge 9\cdot\left[2^{\frac{1}{3}}\cdot (\sqrt{2})^{\frac{1}{3}-4}\right]^{\frac{1}{3}}\quad\left(\text{because}\,\frac{a^2+b^2}{2}\ge ab\right)\\ &\qquad = 9\cdot\left[(\sqrt{2})^{\frac{2}{3}+\frac{1}{3}-4}\right]^{\frac{1}{3}}=9\cdot (\sqrt{2})^{-1}\\ &\qquad =\frac{9\sqrt{2}}{2}. \end{align}$

$\displaystyle\begin{align} LHS &= \frac{1}{\sqrt{2}}\left(\frac{b}{a}+\frac{a}{b}\right)+2\left(\frac{\sqrt{a^2+b^2}}{b}+\frac{2b}{\sqrt{a^2+b^2}}\right)+\frac{1}{\sqrt{2}}\cdot\frac{b}{a}-\frac{2b}{\sqrt{a^2+b^2}}\\ &\ge\frac{1}{\sqrt{2}}\cdot 2+2\cdot 2\sqrt{2}+\frac{1}{\sqrt{2}}\cdot\frac{b}{a}-\frac{2b}{\sqrt{2ab}}\\ &=5\sqrt{2}+\frac{1}{\sqrt{2}}\left(\sqrt{\frac{b}{a}}-1\right)^2-\frac{1}{\sqrt{2}}\\ &\ge\frac{9}{\sqrt{2}}. \end{align}$

For equality, $\displaystyle\frac{a}{b}=\frac{b}{a}\,$ and $\displaystyle\frac{\sqrt{a^2+b^2}}{b}=\frac{2b}{\sqrt{a^2+b^2}},\,$ implying $a=b.$

The inequality is homogeneous which lets us to assume, WLOG, that $a^2+b^2=4.\,$ If so, by the AM-GM inequality, $ab\le 2.\,$ The inequality becomes

$\displaystyle\begin{align} LHS&=\frac{a}{b\sqrt{2}}+\left(\frac{b\sqrt{2}}{2a}+\frac{b\sqrt{2}}{2a}\right)+\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}\right)+\left(\frac{b}{2}+\frac{b}{2}\right)\\ &\overset{AM-GM}{\ge}9\sqrt[9]{ \frac{ab^4\cdot 2}{b^5\cdot a^2\cdot 16\sqrt{2}}}\overset{2\ge ab}{\ge}9\sqrt[9]{\frac{1}{16\sqrt{2}}}\\ &=\frac{9}{\sqrt{2}}. \end{align}$

The problem above has been kindly posted to the CutTheKnotMath facebook page by Dan Sitaru, along with several solutions. Solutions 1 is by Soumava Chakraborty; Solution 2 by Seyran Ibrahimov; Solution 3 is by Su Tanaya; Solution 4 is by Kunihiko Chikaya; Solution 5 is by Imad Zak.