Can we show $x>\ln(1+x^2)$ for all $x>0$?

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We can compute \(\dfrac{dy}{dx}\) using the chain rule: \[\begin{align*} \frac{dy}{dx} &= 1-\frac{1}{1+x^2}\times\frac{d}{dx}(1+x^2)\\ &= 1-\frac{2x}{1+x^2}\\ &= \frac{1+x^2-2x}{1+x^2}\\ &= \frac{(x-1)^2}{1+x^2}. \end{align*}\]

We know that \((x-1)^2 \geq 0\) and \(1+x^2\geq 1\) for all values of \(x\), so \(\dfrac{dy}{dx} \geq 0\) for all values of \(x\).

We know that \(\dfrac{dy}{dx}\geq 0\) for all values of \(x\), so the function \(x-\ln(1+x^2)\) is an increasing function.

Substituting we find that when \(x = 0\), \(y=0\) and \(\dfrac{dy}{dx} = 1\). This means that for any value of \(x>0\), \(y\) must be greater than zero. So \[\begin{align*} x &- \ln(1+x^2) > 0\\ \implies\quad x &> \ln(1+x^2) \end{align*}\]

for all \(x>0\) as required.