The expression whose minimum value I'm trying to find is:
$$\displaystyle\frac{{ \sec{⁴}\alpha}}{{ \tan{²}\beta}}+\frac{{ \sec{⁴}\beta}}{{ \tan{²}\alpha}}$$
I know there are many more methods to find the minimum value using the AMGM inequality. But I want to particularly stress on the following method because it does not give the right answer.
$$\displaystyle\frac{{{\left({1}+{x}\right)}²}}{{{y}}}+\frac{{{\left({1}+{y}\right)}²}}{{{x}}}$$
Where x is tan²α and y is tan²β
$$\displaystyle f{{\left({x},{y}\right)}}=\displaystyle\frac{{{x}²}}{{y}}+\frac{{{y}²}}{{x}}+\frac{{{2}{y}}}{{x}}+\frac{{{2}{x}}}{{y}}+\frac{1}{{x}}+\frac{1}{{y}}$$
Now let this be f(x,y)
Apply AM GM inequality on all terms of f(x,y) to get
$$\displaystyle{A}{M}\ge{G}{M}$$
Apply this on the terms of f(x,y)
$$\displaystyle\frac{{ f{{\left({x},{y}\right)}}}}{{6}}\ge{\left[\frac{{{x}²}}{{y}}\cdot\frac{{{y}²}}{{x}}\cdot\frac{1}{{x}}\cdot\frac{1}{{y}}\cdot\frac{{{2}{x}}}{{y}}\cdot\frac{{{2}{y}}}{{x}}\right]}^{{\frac{1}{{6}}}}$$
Simplify the RHS to get
$$\displaystyle\frac{ f{{\left({x},{y}\right)}}}{{6}}\ge{4}^{{\frac{1}{{6}}}}$$
Simplify to get
f(x,y) is greater than or equal to $$\displaystyle f{{\left({x},{y}\right)}}\ge{6}\cdot{2}^{{\frac{1}{{3}}}}$$
THIS NUMBER IN THE RHS IS APPROXIMATELY 7.55
why on earth do we get it's minimum value as 7.55 when we use this method, when the actual minimum value is 8. AMGM inequality is supposed to give the sharp minima.
Thank you!
Best Answer
To summarize the discussion in the comments:
The problem here is that, by dividing into $6$ asymmetric terms, you have made it impossible for all the terms to be equal, so you can't realize the inequality sharply.
To illustrate the problem, consider the usual AM-GM example: $$x>0\implies x+\frac 1x≥2$$ as always, that bound is realized when the two terms are equal (hence when $x=1$). Now suppose you split the left hand into four terms $$\frac x3+\frac {2x}3+\frac {1}{2x}+\frac 1{2x}$$ Now the terms can't be equal (as the second is always twice the first). You can still apply AM-GM of course and now you get $$x+\frac 1x=\frac x3+\frac {2x}3+\frac {1}{2x}+\frac 1{2x}≥4\sqrt[4]{\frac 1{18}}=1.9419$$
Perfectly true, but not sharp.