Proving inequalities via the mean value theorem

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Prove the following inequalities:

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(a)

∣sin⁡x−sin⁡y∣⩽∣x−y∣\left|\sin x-\sin y\right| \leqslant \left|x-y\right|;

(b)

if 0<y<x0<y<x and p>1p>1, then pyp−1(x−y)<xp−yp<pxp−1(x−y)py^{p-1}(x-y)<x^{p}-y^{p}<px^{p-1}(x-y);

(c)

∣arctg⁡a−arctg⁡b∣⩽∣a−b∣\left|\arctg a-\arctg b\right| \leqslant \left|a-b\right|;

(d)

a−ba<ln⁡ab<a−bb\dfrac {a-b}{a}<\ln \dfrac {a}{b}<\dfrac {a-b}{b}, where 0<b<a0<b<a.

Using Lagrange's formula, prove the inequality n(b−a)an−1<bn−an<n(b−a)bn−1n(b-a)a^{n-1} < b^{n}-a^{n} < n(b-a)b^{n-1} for b>a>0b > a > 0, n>1n > 1.

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Using Lagrange's formula, prove the inequality x1+x<ln⁡(1+x)<x\dfrac {x}{1+x} < \ln (1+x) < x for x>0x > 0.

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Using Lagrange's formula, prove the inequality ex>1+xe^{x} > 1+x.

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Using Lagrange's formula, prove the inequality ex>exe^{x} > ex for x>1x > 1.

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Using Lagrange's formula, prove the inequality xδ∣ln⁡x∣<1δex^{\delta }\left|\ln x\right| < \dfrac {1}{\delta e} for 0<x<10 < x < 1, δ>0\delta > 0.

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Prove the theorem: if at x=0x=0 the functions φ(x),φ′(x),…,φ(n−1)(x)\varphi (x), \varphi '(x), \ldots , \varphi^{(n-1)}(x) vanish, and the function φ(n)(n)\varphi^{(n)}(n) is positive for x>0x > 0, then φ(x)>0\varphi (x) > 0 for positive xx.

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Prove the theorem: if

φ(0)=ψ(0),  φ′(0)=ψ′(0),  …,  φ(n−1)(0)=ψ(n−1)(0) \varphi (0) = \psi (0), \; \varphi '(0) = \psi '(0), \; \ldots , \; \varphi ^{(n-1)}(0) = \psi ^{(n-1)}(0)

and φ(n)(x)>ψ(n)(x)\varphi^{(n)}(x) > \psi^{(n)}(x) for x>0x > 0, then for positive xx the inequality φ(x)>ψ(x)\varphi (x) > \psi (x) holds.

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Prove the inequality x−x36<sin⁡x<xx-\dfrac {x^{3}}{6} < \sin x < x, for x>0x > 0.

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Prove the inequality x−x33<arctg⁡x<xx-\dfrac {x^{3}}{3} < \arctg x < x, for x>0x > 0.

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Prove the inequality tg⁡x>x+x33\tg x > x+\dfrac {x^{3}}{3}, for 0<x<π20 < x < \dfrac {\pi }{2}.

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Prove the inequality x−x22<ln⁡(1+x)<xx-\dfrac {x^{2}}{2} < \ln (1+x) < x; x>0x > 0.

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By direct squaring, prove the inequalities

1+x2−x28<1+x<1+x2for 0<x<8. 1+\frac{x}{2}-\frac{x^{2}}{8} < \sqrt{1+x} < 1+\frac{x}{2} \quad \text{for } 0 < x < 8.
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Prove that for 0<x<10 < x < 1 the quantity e2xe^{2x} is less than 1+x1−x\dfrac {1+x}{1-x}.

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Applying Lagrange's theorem to the function 1/xα1 / x^{\alpha }, prove that for any n∈Nn \in \mathbb {N} and any α>0\alpha > 0 the inequality holds

1nα+1<1α[1(n−1)α−1nα]. \frac{1}{n^{\alpha +1}} < \frac{1}{\alpha }\left[\frac{1}{(n-1)^{\alpha }}-\frac{1}{n^{\alpha }}\right].
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Using Lagrange's theorem, prove the inequalities:

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(1)

n(b−a)an−1<bn−an<n(b−a)bn−1n(b-a) a^{n-1} < b^{n}-a^{n} < n(b-a) b^{n-1} for 0<a<b,n∈N0 < a < b, n \in \mathbb {N};

(2)

x/(1+x)<ln⁡(1+x)<xx /(1+x) < \ln (1+x) < x for x>0x > 0;

(3)

ex⩾1+xe^{x} \geqslant 1+x for x∈Rx \in \mathbb {R};

(4)

ex>exe^{x} > e x for x>1x > 1;

(5)

xα∣ln⁡x∣,1/(αe)x^{\alpha }\left|\ln x\right|, 1 /(\alpha e) for 0<x<1,α>00 < x < 1, \alpha > 0.

Prove: if the function ff is differentiable nn times for x⩾0x \geqslant 0, f(0)=f′(0)=…=f(n−1)(0)=0f(0) = f^{\prime }(0) = \ldots = f^{(n-1)}(0) = 0, and f(n)(x)>0f^{(n)}(x) > 0 for x>0x > 0, then also f(x)>0f(x) > 0 for x>0x > 0.

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Prove Theorem 2.4.20.

Theorem 2.4.20: Let ff be a differentiable function on a set E⊂RE \subset \mathbb {R}. Then ff is Lipschitz with constant K=sup⁡{∣f′(x)∣:x∈E}K = \sup \left\{ \left|f'(x)\right| : x \in E\right\} if and only if it has a bounded derivative.

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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Предел. Непрерывность. Дифференцируемость. 2-е изд., ФИЗМАТЛИТ, 2003.
[Pet25] Petrovic J.S. Measure and Integral: Theory and Practice. CRC Press, 2025.