Articles of inequality

with inequality $\frac{1}{3a+5b+7c}+\frac{1}{3b+5c+7a}+\frac{1}{3c+5a+7b}\le\frac{\sqrt{3}}{4}$

let $a,b,c>0$, such $ab+bc+ac=1$,show that $$\dfrac{1}{3a+5b+7c}+\dfrac{1}{3b+5c+7a}+\dfrac{1}{3c+5a+7b}\le\dfrac{\sqrt{3}}{4}$$ by Macavity C-S:with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$ $$\dfrac{1}{3a+5b+7a}\le\dfrac{1}{2}\left(\dfrac{1}{6a+9b}+\dfrac{1}{b+7c}\right)$$ It suffices to show $$\sum_{cyc}\left(\dfrac{1}{6a+9b}+\dfrac{1}{b+7c}\right)\le\dfrac{\sqrt{3}}{2}$$

Will an increasing function applied inside definite integrals preserve inequality?

If $$\int_a^b f(x) \,dx > \int_a^b g(x) \,dx$$ and there is a function $h(x)$ that is strictly increasing with $x$, does that imply that $$\int_a^b h(f(x)) \,dx > \int_a^b h(g(x)) \,dx$$ ?

Proof of inequality without calculus

So we are given the equation $3x+4y+xy=2012$ where $x$ and $y$ are positive integers. Prove that $x+y\geq83$. Using calculus optimisation methods, this can be proved. However, it requires a lot of difficult calculations, and students are not expected to know calculus to answer this. How else could you prove this? You could always go through […]

For $x>0$, $x + \frac1x \ge 2$ and equality holds if and only if $x=1$

This question already has an answer here: How to prove this inequality $ x + \frac{1}{x} \geq 2 $ 17 answers

Proof of $n^{1/n} – 1 \le \sqrt{\frac 2n}$ by induction using binomial formula

This question already has an answer here: Prove that $\,\sqrt [n] n < 1 + \sqrt{\frac{2}{n}}\,$ 1 answer

A qualifying exam problem involving Schwarz lemma

This is a problem in the book “Berkeley Problems in Mathematics”, which I think the solution given is wrong, can someone help? The following problem appeared in Spring 1991. Let the function $f$ be analytic in the unit disc, with $|f(z)|\leqslant 1$ and $f(0)=0$. Assume that there is a number $r$ in $(0,1)$ such that […]

Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$

Let $a,b,$ and $c$ be positive real numbers with $ab+bc+ca = 1$. Prove that $$\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{3}{2}$$ Attempt The $ab+bc+ca = 1$ condition reminds of the rearrangement inequality. Thus, I would say that $a^2+b^2+c^2 \geq ab+bc+ca = 1$ then rewrite the given inequality as $4(a+b+c)^2 = 4(a^2+b^2+c^2) + 8(ab+bc+ca) = 4(a^2+b^2+c^2) \leq 9(a^2+1)(b^2+1)(c^2+1)$ I don’t […]

Inequality understanding

My textbook says that: $$ \frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)^2\cdots\left(1+\frac{1}{n}\right)^n<e^n $$ But I do not understand this. Can you please enlighten me? Edit: How do you show that $$ \frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)^2\cdots\left(1+\frac{1}{n}\right)^n $$

A combinatorial inequality

How can I prove $$ \log \binom nk \leq k \left(1 +\log\frac{n}{k}\right) $$ where $\binom\cdot\cdot$ stands for combination. I tried to use stirling approximation but I couldn’t get the inequality.

How prove this inequality $(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$

let $a,b,c$ are postive numbers, show that $$(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$$ my try: let $$a+b+c=p,ab+bc+ac=q,abc=r$$ and the $$a^3+b^3+c^3=(a+b+c)^3-3(ab+bc+ac)(a+b+c)=p^3-3pq$$ Thank you