Problems

Is it true that if $b$ is a positive number, then $b^3+b^2\ge b$? What about $b^3+1\ge b$?

Show that if $a$ is positive, then $1+a\ge 2\sqrt{a}$.

Let $k$ be a natural number, prove the following inequality. $$\frac{1}{k^2}>\frac{1}{k}-\frac{1}{k+1}.$$

Show that if $a$ is a positive number, then $a^3+2\ge 2a\sqrt{a}$.

The numbers $a$, $b$ and $c$ are positive. By completing the square, show that $$\frac{a^2}{4}+b^2+c^2\ge ab-ac+2bc.$$

Let $m$ and $n$ be natural numbers such that $m>n$. Show that: $$\frac{1}{n^2}+\frac{1}{(n+1{)}^2}+\frac{1}{(n+2{)}^2}+\cdots +\frac{1}{m^2}>\frac{1}{n}-\frac{1}{m}.$$

The numbers $a,b,c$ are positive. Show that: $$\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c.$$

The number $n$ is natural. Show that: $$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{n}}<3\sqrt{n+1}-3.$$

If $n$ is a positive integer, we denote by $s(n)$ the sum of the divisors of $n$. For example, the divisors of $n=6$ are $1,2,3,6$, so $s(6)=1+2+3+6=12$. Prove that, for all $n\ge 1$, $$s(1)+s(2)+\cdots +s(n)\le n^2.$$ Denote by $t(n)$ is instead the sum of the squares of the divisors of $n$ (e.g., $t(6)=1^2+2^2+3^2+6^2=50$), can you find a similar inequality for $t(n)$?

Consider the following sum: $$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\dots$$ Show that no matter how many terms it has, the sum will never be larger than $1$.