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 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 […]

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.

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 […]

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: $$n=2$$ $$(1-x)^2 + (1+x)^2 < 2^2$$ $$(1-2x+x^2) + (1+2x+x^2) < 2^2$$ $$2+2x^2 < 2^2$$ $$2(1+x^2) < 2^2$$ […]

Let $n$ points be given in the unit square. How to prove or disprove: the points can be labeled $x_1,\ldots,x_n$ to satisfy the inequality $$\|x_1-x_2\|^2 +\|x_2-x_3\|^2+\cdots+\|x_n-x_1\| ^2 \le 4,$$ where $\|\cdot\| $ is the Euclidian distance?

Can anyone prove the Ptolemy inequality, which states that for any convex quadrulateral $ABCD$, the following holds:$$\overline{AB}\cdot \overline{CD}+\overline{BC}\cdot \overline{DA} \ge \overline{AC}\cdot \overline{BD}$$ I know this is a generalization of Ptolemy’s theorem, whose proof I know. But I have no idea on this one, can anyone help?

This is just a standard problem from my high school’s calculus text, but my proof seems sort of off. This is it: Let $f(x) = e^x$. The tangent line of $f(x)$ at $x=0$ is $g(x)=x+1$. Since $f”(x_0) \gt 0$ for all $x_0 \in \mathbb R$,the tangent line $g(x) \le f(x)$ for all $x$. Q.E.D. I […]

Inequality Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality $$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$ I stumbled upon this question some days ago and been trying AM-GM to find the solution but so far have been unsuccessful.

How can I prove the following inequality: Given $ a,b>0 $ and $a^2>b $, we have $a>\sqrt b$ Thank you.

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