Articles of inequality

$a,b,c$ are positive reals and distinct with $a^2+b^2 -ab=c^2$. Prove $(a-c)(b-c)<0$

This question already has an answer here: If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$. 6 answers

How to find the minimum value of this function?

How to find the minimum value of $$\frac{x}{3y^2+3z^2+3yz+1}+\frac{y}{3x^2+3z^2+3xz+1}+\frac{z}{3x^2+3y^2+3xy+1}$$,where $x,y,z\geq 0$ and $x+y+z=1$. It seems to be hard if we use calculus methods. Are there another method? I have no idea. Thank you.

Given that $abc=1$, prove that $\frac{b}{c} + \frac{c}{a}+ \frac{a}{b} \ge \frac{1}{a} + \frac{1}{b} +\frac{1}{c}$.

I have tried to do a bunch of different versions of AM-GM on this and the equivalent versions that come from substituting in $abc=1$ and I always flip the inequality by doing it. Help is greatly appreciated.

Proving $a^ab^bc^c\ge(abc)^{(a+b+c)/3}$ for positive real numbers.

This question already has an answer here: Prove $a^ab^bc^c\ge (abc)^{\frac{a+b+c}3}$ for positive numbers. 3 answers

Pure algebra: Show that this expression is positive

Suppose that $n\geq2$, $g\in(0,1)$ and $s\in(0,1)$. Show that: $1-(1-g)ns(1-(1-g)s)^{n-1}-(1-(1-g)s)^n > 0 $ What I have done: I have confirmed that given my assumptions this inequality holds with the help of Mathematica. Here a picture of how the function looks like. In the y-axis the value of the function and in the x-axis the value of […]

Inequality $a^2+b^2+c^2 \leq a^2b+b^2c+c^2a+1. $

Prove that: $$a^2+b^2+c^2 \leq a^2b+b^2c+c^2a+1, (\forall) a,b,c \in [0,1].$$ I have no idea, I try $AM\geq GM$ but still nothing.

How would I prove $|x + y| \le |x| + |y|$?

How would I write a detailed structured proof for: for all real numbers $x$ and $y$, $|x + y| \le |x| + |y|$ I’m planning on breaking it up into four cases, where both $x,y < 0$, $x \ge 0$ and $y<0$, $x<0$ and $y \ge0$, and $x,y \ge 0$. But I’m not sure how […]

Show that for any integer $n\ge 6$, the inequality $2^n>7n$ holds.

Step one: for case where $n=6$ $$7n <2^n$$ $$7(6)<2^6 \rightarrow 42<64$$. Step two: Suppose for $n$ such that $7n<2^n$ is true. Now prove for $n+1$ $$(2)7n<(2)2^n$$ $$14n<2^{n+1}$$ But since $n<7n<2^n$, then $n+7<7(n+1)<2^n+7<2^{n+1}(?)$. Then $7(n+1)<2^{n+1}$ I’m new in this induction process, so any help/tips for this problem would be really appreciated.

Inequality $\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$

If $x>0$, $y>0$, $z>0$ and $xyz = 1$ then $$\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2} \le \sqrt{2}(x+y+z)$$ I tried using $\displaystyle x = \frac{a}{b},y = \frac{b}{c}$ and $\displaystyle z = \frac{c}{a}$ substitution, $\displaystyle \sum_{cyc} \frac{\sqrt{a^2+b^2}}{b} \le \sqrt{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \\ \displaystyle \iff 6+ 2\sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le \sum_{cyc}\frac{a^2+b^2}{b^2} + 4\sum_{cyc}\frac{b}{a}$ So it would suffice if we showed $\displaystyle \sum_{cyc}\frac{\sqrt{(a^2+b^2)(b^2+c^2)}}{bc} \le 2\sum_{cyc}\frac{b}{a}$ Squaring again does […]

Prove an inequality by Induction: $(1-x)^n + (1+x)^n < 2^n$

Could you give me some hints, please, to the following problem. Given $x \in \mathbb{R}$ such that $|x| < 1$. Prove by induction the following inequality for all $n \geq 2$: $$(1-x)^n + (1+x)^n < 2^n$$ $1$ Basis: $$n=2$$ $$(1-x)^2 + (1+x)^2 < 2^2$$ $$(1-2x+x^2) + (1+2x+x^2) < 2^2$$ $$2+2x^2 < 2^2$$ $$2(1+x^2) < 2^2$$ […]