An inequality in 2+2 variables from SSMA magazine


An inequality in 2+2 variables from SSMA magazine

Solution 1

We may assume that both $k,l$ are positive because in the right-hand side $\displaystyle \frac{2|k||l|}{\sin (\alpha+\beta)}\ge\frac{2kl}{\sin (\alpha+\beta)},$ and this is the only term that is affected by the change of the signs. Thus

$\displaystyle \begin{align} &k^2\cos^2 \beta +l^2\cos^2 \alpha\overbrace{\geq}^{AM-GM}2\sqrt{k^2l^2\cos^2\alpha \cos^2 \beta}\geq 2kl \cos \alpha \cos \beta\\ &k^2\cos^2 \beta+l^2\cos^2 \alpha -2kl \cos \alpha \cos \beta \geq 0\\ &\frac{k^2\cos^2 \beta}{\sin (\alpha+\beta)\cos \alpha \cos \beta}+\frac{l^2\cos^2 \alpha}{\sin (\alpha+\beta)\cos \alpha \cos \beta}-\frac{2kl \cos \alpha \cos \beta}{\sin (\alpha+\beta)\cos \alpha \cos \beta}\geq 0\\ &\frac{k^2\cos \beta}{\sin (\alpha+\beta)\cos \alpha}+\frac{l^2 \cos \alpha}{\sin (\alpha+\beta)\cos \beta}-\frac{2kl}{\sin (\alpha+\beta)}\geq 0\\ &\frac{k^2\cos (\alpha+\beta-\alpha)}{\sin (\alpha+\beta)\cos \alpha}+\frac{l^2\cos (\alpha+\beta -\beta)}{\sin (\alpha+\beta)\cos \beta}-\frac{2kl}{\sin (\alpha+\beta)}\geq 0\\ &k^2\Bigr(\frac{\sin \alpha}{\cos \alpha}+\frac{\cos (\alpha+\beta)}{\sin (\alpha+\beta)}\Bigr)+l^2 \Bigr(\frac{\sin \beta}{\cos \beta}+\frac{\cos (\alpha+\beta)}{\sin (\alpha+\beta)}-\frac{2kl}{\sin (\alpha+\beta)}\geq 0\\ &k^2 \Bigr(\tan \alpha+\cot (\alpha+\beta)\Bigr)+l^2\Bigr(\tan \beta+\cot (\alpha+\beta)\Bigr)\geq \frac{2kl}{\sin (\alpha+\beta)}\\ &k^2 \tan \alpha+l^2 \tan \beta \geq \frac{2kl}{\sin (\alpha+\beta)}-(k^2+l^2)\cot (\alpha+\beta) \end{align}$

Solution 2

We rewrite the inequality by transposing to obtain a sequence of equivalent inequalities:

  • $\displaystyle k^2\left(\frac{\sin\alpha}{\cos\alpha}+\frac{\cos(\alpha+\beta)}{\sin(\alpha+\beta)}\right)+l^2\left(\frac{\sin\beta}{\cos\beta}+\frac{\cos(\alpha+\beta)}{\sin(\alpha+\beta)}\right)\ge\frac{2kl}{\sin(\alpha+\beta)}$

  • Multiplying by $\sin(\alpha+\beta),$

  • $\displaystyle\begin{align}&k^2\left(\frac{\sin\alpha\cdot\sin(\alpha+\beta)+\cos\alpha\cdot\cos(\alpha+\beta)}{\cos\alpha}\right)\\ &\qquad\qquad\qquad+l^2\left(\frac{\sin\beta\cdot\sin(\alpha+\beta)+\cos\beta\cdot\cos(\alpha+\beta)}{\cos\beta}\right)\ge 2kl\end{align}$

  • $\displaystyle k^2\cdot\frac{\cos\beta}{\cos\alpha}+l^2\cdot\frac{\cos\alpha}{\cos\beta}\ge 2kl$

  • $\displaystyle \frac{k^2\cos^2\beta+l^2\cos^2\alpha}{\cos\alpha\cos\beta}\ge 2kl$

  • $\displaystyle k^2\cos^2\beta+l^2\cos^2\alpha\ge 2kl\cos\alpha\cos\beta,$ and, by transposing,

  • $(k\cos\beta-l\cos\alpha)^2\ge 0.$

  • So we retrace our steps to obtain the original inequality.

Solution 3

$\displaystyle k^2 \tan (\alpha )+l^2 \tan (\beta )\geq \frac{2 k l}{\sin (\alpha +\beta )}-\left(k^2+l^2\right) \cot (\alpha +\beta )$

can be rewritten as:

$\displaystyle \sec (\alpha ) \sec (\beta ) \csc (\alpha +\beta ) \left(l \cos (\alpha )-k \cos (\beta )\right)^2\geq 0$

which proves the inequality.


Dan Sitaru has kindly communicated to me the above problem, along with a solution of his (Solution 1). The problem was proposed by Arcady Alt. Solution 2, by Ed Gray, has been published by the magazine. Solution 3 is by N. N. Taleb.


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