Exercise 1268Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL1Added 4 Nov 2026

An arctangent inequality

#mean value theorem#inequality#arctan

Statement

Prove that for all $t>0$ : $\arctan t>\frac{t}{1+t^2}$ .

Hint

Apply the MVT to $\arctan$ on $[0,t]$ ; bound $\frac{1}{1+c^2}$ using $0<c<t$ .

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MVT. There is $c\in]0,t[$ with $\arctan t=\frac{t}{1+c^2}$ .

Bound. Since $c<t$ , $1+c^2<1+t^2$ , so $\frac{1}{1+c^2}>\frac{1}{1+t^2}$ . Multiply by $t>0$ :
$\arctan t>\frac{t}{1+t^2}.$

Conclusion. The inequality holds.