Let $f$ be a continuous and differentiable function in $(a,b)$. If $f(x)f'(x) \ge x\sqrt{1 - (f(x))^4}$ and $\lim_{x\to a^+}(f(x))^2 = 1$ and $\lim_{x\to b^-} (f(x))^2 = \frac 12$; for $x \in (a,b)$; then prove that $a^2 - b^2 \ge \frac \pi3$. Assume that $\lim_{x\to a} f(g(x) = f(\lim_{x\to a}g(x))$ holds true.
Application of Derivatives: $f(x)f'(x) \ge x\sqrt{1 - (f(x))^4}$
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calculus
limits
derivatives
1 Answers
3
Let $y=(f(x)) ^{2}$ then we can see that $$\int_{a}^{b} \frac{f(x) f'(x)} {\sqrt{1-(f(x))^{4}}}=-\frac{1}{2}\int_{1/2}^{1}\frac{dy}{\sqrt{1-y^{2}}}=-\frac{\pi} {6}$$ and this is greater than or equal to $\int_{a} ^{b} x\, dx$ and we are done.