Da: $$f(\tan2x)=\tan^{4}x+\frac{1}{\tan^{4}x}, \forall x\in\left(0;\frac{\pi}{4}\right)$ $
Demostrar que: $f(\sin x)+f(\cos x) \ge 196, \forall x\in\left(0;\frac{\pi}{2}\right)$
¿Alguien me podria ayudar?
Respuesta
¿Demasiados anuncios?
desde $$t=\tan{2x}=\dfrac{2\tan{x}}{1-\tan^2{x}}=-\dfrac{2}{\tan{x}-\dfrac{1}{\tan{x}}},t>0$ $ % que $$\tan{x}-\dfrac{1}{\tan{x}}=-\dfrac{2}{t}$$ y $$\tan^4{x}+\dfrac{1}{\tan^4{x}}=\left[\left(\tan{x}-\dfrac{1}{\tan{x}}\right)^2+2\right]^2-2$ $ lo $ de $$f(t)=\left[\dfrac{4}{t^2}+2\right]^2-2=\dfrac{16}{t^4}+\dfrac{16}{t^2}+2$ $$f(\sin{x})+f(\cos{x})=16\left(\dfrac{1}{\sin^4{x}}+\dfrac{1}{\cos^4{x}}\right)+16\left(\dfrac{1}{\sin^2{x}}+\dfrac{1}{\cos^2{x}}\right)+4$ $ desde utilizar desigualdad titular tenemos $$\left(\dfrac{1}{\sin^4{x}}+\dfrac{1}{\cos^4{x}}\right)(\sin^2{x}+\cos^2{x})^2\ge (1+1)^3$ $ $$\Longrightarrow \dfrac{1}{\sin^4{x}}+\dfrac{1}{\cos^4{x}}\ge 8$ $ $$\left(\dfrac{1}{\sin^2{x}}+\dfrac{1}{\cos^2{x}}\right)(\sin^2{x}+\cos^2{x})\ge (1+1)^2$ $ % $ $$\Longrightarrow \dfrac{1}{\sin^2{x}}+\dfrac{1}{\cos^2{x}}\ge 4$ que $$f(\sin{x})+f(\cos{x})\ge 196$$
