Articles of inequality

A natural proof of the Cauchy-Schwarz inequality

Most of the proofs of the Cauchy-Schwarz inequality on a pre-Hilbert space use a fact that if a quadratic polynomial with real coefficients takes positive values everywhere on the real line, then its discriminant is negative(e.g. Conway: A course in functional analysis). I think this is somewhat tricky. Moreover I often forget its proof when […]

A (probably trivial) induction problem: $\sum_2^nk^{-2}\lt1$

So I’m a bit stuck on the following problem I’m attempting to solve. Essentially, I’m required to prove that $\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} < 1$ for all $n$. I’ve been toiling with some algebraic gymnastics for a while now, but I can’t seem to get the proof right. Proving it using calculus isn’t a problem, but I’m struggling […]

Prove that $d(n)\leq 2\sqrt{n}$

In Waclaw Sierpinski’s book Elementary Theory of Numbers on page 168 there is the following exercise: “Exercises. 1. Prove that for natural numbers $n$ we have $d(n) \leq 2\sqrt{n}$,” where $d(n)$ is the number of divisors of n. As a hint right below is given: “The proof follows from the fact that of two complementary […]

How to prove these inequalities: $\liminf(a_n + b_n) \leq \liminf(a_n) + \limsup(b_n) \leq \limsup(a_n + b_n)$

This question already has an answer here: Prove $\limsup\limits_{n \to \infty} (a_n+b_n) \le \limsup\limits_{n \to \infty} a_n + \limsup\limits_{n \to \infty} b_n$ 4 answers

The series $\sum\limits_{n=1}^\infty \frac n{\frac1{a_1}+\frac1{a_2}+\dotsb+\frac1{a_n}}$ is convergent

If a series $\sum\limits_{n=1}^\infty a_n$ is convergent, and $a_n\gt0$… Do not refer to Carleman’s inequality or Hardy’s inequality, show that the series $$\sum_{n=1}^\infty \frac n{\frac1{a_1}+\frac1{a_2}+\dotsb+\frac1{a_n}} $$ is also convergent. What is the minimum positive real number $k$ such that the following inequality holds for all convergent series $a_n\gt0$? $$\sum_{n=1}^\infty \frac n{\frac1{a_1}+\frac1{a_2}+\dotsb+\frac1{a_n}}\le k\sum_{n=1}^\infty a_n$$ Does it […]

Prove : $\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$

If we assume that: $0\le x_1,x_2,\ldots,x_{10}\le\frac{\pi}{2} $ such that: $$\sin^2 (x_1) +\sin^2 (x_2)+\cdots+\sin^2(x_{10})=1$$ How to prove that: $$\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$$

$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$

$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $$x^2y+y^2z+z^2x < \frac12$$ This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.

Show that $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} \geq x+y+z $ by considering homogeneity

Well, I’m preparing for an undergrad competition that is held in April and because of that I’ve been trying to solve the inequalities I find on the internet. I found this problem: $$\displaystyle \frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} \geq x+y+z \text{ for all } x,y,z \in \mathbb{R^+}$$ It’s easy to show that this inequality holds […]

Prove that in an ordered field $(1+x)^n \ge 1 + nx + \frac{n(n-1)}{2}x^2$ for $x \ge 0$

In an ordered field show that $x \geq 0 \implies (1+x)^{n} \geq 1+nx+ \frac{1}{2}n(n-1)x^2$ for every positive integer $n$. I know that $(1+x)^{n} \geq 1+nx$ (Bernoulli’s inequality). To get the stronger inequality you can probably use induction also. But is there an easier way? The extra term on the right seems to be an “error” […]

proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$

If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ and also show that equality hold if $a=b=c$. $\bf{My\; Try}::$ Here we have to prove $4\triangle\leq \sqrt{(a+b+c)\cdot abc}$ Using the formula $$\triangle = \sqrt{s(s-a)(s-b)(s-c)},$$ where $$2s=(a+b+c)$$ So $$4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$ Now using $\bf{A.M\geq G.M}$ for $(b+c-a)\;,(c+a-b)\;,(a+b-c)>0$ […]