0

If $$ \frac{\cos^4(x)}{\cos^2(y)} + \frac{\sin^4(x)}{\sin^2(y)} = 1$$ To find the value of : $$ \frac{\cos^4(y)}{\cos^2(x)} + \frac{\sin^4(y)}{\sin^2(x)} $$ using Titu's Inequality

So far this is my progress:
By Titu's Inequality

$$ \frac{(\cos^2(x))^2}{\cos^2(y)} + \frac{(\sin^2(x))^2}{\sin^2(y)} \ge \frac{(\cos^2(x))^2 + (\sin^2(x))^2}{\cos^2(y) + \sin^2(y)} \ge 1$$

Since there's an equality case we can conclude
$$\frac{\cos^2(x)}{\cos^2(y)} = \frac{\sin^2(x)}{\sin^2(y)}$$ or $$\frac{\cos^2(y)}{\cos^2(x)} = \frac{\sin^2(y)}{\sin^2(x)}$$

But I find myself stuck from this point in finding the value of the required.
Thanks

Robert Z
  • 145,942
AYUSH BANERJEE
  • 1
  • the part of inequality mentioned $$ \frac{(\cos^2(x))^2 + (\sin^2(x))^2}{\cos^2(y) + \sin^2(y)} \ge 1$$ doesnt necessarily hold because by Titu's Lemma we have $$ \frac{(\cos^2(x) + \sin^2(x))^2}{\cos^2(y) + \sin^2(y)} \ge 1$$ so we have $$ {(\cos^2(x) + \sin^2(x))^2} =\cos^4(x) + \sin^4(x) + 2\cos^2(x)\sin^2(x) $$ so $$2\cos^2(x)\sin^2(x) \ge 0 $$ so this holds $$ \frac{(\cos^2(x))^2 + (\sin^2(x))^2}{\cos^2(y) + \sin^2(y)} \ge 1- 2\frac{\cos^2(x)*\sin^2(x)}{\cos^2(y) + \sin^2(y)}$$ can you elaborate how did you obtain $$\frac{\cos^2(x)}{\cos^2(y)} = \frac{\sin^2(x)}{\sin^2(y)}$$? – thomas graceman Jul 28 '23 at 11:57
  • Well I ain't good with LaTeX but uhh
    $(sin^2x + cos^2x)^2/(sin^2y + cos^2y) = 1$ and we're given $sin^4x/sin^2y + cos^4x/cos^2y = 1$
    we conclude $(sin^2x + cos^2x)^2/(sin^2y + cos^2y) = sin^4x/sin^2y + cos^4x/cos^2y $

    Such equality holds in Titu's iff $sin^4x/sin^2y = cos^4x/cos^2y$ i.e. $(sin^2x)^2/sin^2y = (cos^2x)^2/cos^2y$

    From here I'm stuck again but I feel it'll reduce to $sin^2x = sin^2y$

     – AYUSH BANERJEE Jul 28 '23 at 15:28
  • Using Titu's lemma (i.e. CS inequality), I would approach the first part like this: $$1=\frac{\cos^4x}{\cos^2y}+\frac{\sin^4x}{\sin^2y}\geqslant \frac{(\cos^2x + \sin^2x)^2}{\cos^2y+\sin^2y}=1$$ Hence there is equality, implying $\displaystyle \frac{\cos^2x}{\cos^2y}=\frac{\sin^2x}{\sin^2y} = t,$ say.
    Then it isn't difficult to show $t=1$ using the condition and use it to show the expression in question values also $1$.     – Macavity Jul 29 '23 at 02:37
  • @Gonçalo Did you see a linked topic? Did you see a starting problem? If no, do it please. – Michael Rozenberg Jul 29 '23 at 05:18
  • @Gonçalo I said about your previous comment and I wait your reaction. – Michael Rozenberg Jul 29 '23 at 07:27
  • @MichaelRozenberg, replace $(x,y)$ with $(\alpha,\beta)$ and you have the same problem. – Gonçalo Jul 29 '23 at 07:52
  • @Gonçalo Just really no! Read please again, what a topic starter looks out. – Michael Rozenberg Jul 29 '23 at 08:00
  • @Gonçalo What do you think? Can your link help? – Michael Rozenberg Jul 29 '23 at 11:00
  • @Gonçalo the post is indeed helpful, though the reduction of the starting problem there uses manipulations while I started with Titu's Inequality (which probably works as a mere shortcut) since both methods converge to he same logic at the last step i.e to realise one can write sin^2x as 1 - cos^2x, right? – AYUSH BANERJEE Jul 30 '23 at 06:12

1 Answers1

0

Now, $$\frac{\cos^2x}{\cos^2y}=\frac{1-\cos^2x}{1-\cos^2y}$$ or $$\cos^2x-\cos^2x\cos^2y=\cos^2y-\cos^2x\cos^2y$$ or $$\cos^2x=\cos^2y.$$ Thus, $$\frac{\cos^4y}{\cos^2x}+\frac{\sin^4y}{\sin^2x}=\cos^2x+\frac{(1-\cos^2x)^2}{\sin^2x}=\cos^2x+\sin^2x=1.$$

Michael Rozenberg
  • 194,933