I tried proving this with triangular inequality but i was not right can any one help me with this
8964
I've seen the full proof of the Triangle Inequality  \begin{equation*} |x+y|\le|x|+|y|.  \end{equation*} However, I haven't seen the proof of the reverse triangle inequality:  \begin{equation*} ||x|-|y||\le|x-y|. \end{equation*} Would you please prove this using only the Triangle Inequality above?  Thank you very much.
93275
I have looked everywhere for confirmation of this proof of the triangle inequality with no success.  Prove the triangle inequality:  $$\vert x + y \vert \leq \vert x \vert + \vert y \vert.$$  Proof:  Given the definition of $|x|$ to be $|x| = \max\{-x, x\}$. We know that $x \leq \vert x \vert$ and $y \leq \vert y \vert$.  So:  $$\vert x + y \vert \leq \vert \vert x \vert + \vert y \vert \vert = \vert x \vert + \vert y \vert.$$  Thus, we have proven the triangle inequality:  $$\vert x + y \vert \leq \vert x \vert + \vert y \vert.$$  What is wrong with it? If nothing is wrong with this proof, why is it not readily used as it seems to be quite concise.
60631
I tried to solve this inequality because I find exotic.  Actually, I didn't look at the right solution.  Because before I look at the right solution, I want to know if my solution is right or not.  Prove that $0 ≤ yz + zx +xy -2xyz≤\frac{7}{27}$, where $x,y$ and $z$ are non-negative real numbers for which $x + y + z = 1.$  Attempts: If $x=0$. The left side of the inequality is correct. So, I can accept $x,y,z≠0$. It is enough to prove $\frac 1x+\frac1y+\frac1z≥2$ We have,  $x+y+z≥3\sqrt[3]{xyz}\Longrightarrow xyz≤\frac1{27}$ $\frac 1x+\frac1y+\frac1z≥\frac3{\sqrt[3]{xyz}}≥9≥2$ The left side proved.  It is obvious at least one of the numbers is less than $\frac 12.$  So, we can choose $y$, such that $y≤\frac 12$.  Therefore, we have  $yz + zx +xy −2xyz ≤ \frac7{27}$ $x(y+z)+yz(1-2x)-\frac7{27}≤0$ $x(1-x)+y(1-x-y)(1-2x)-\frac7{27}≤0$ $x-x^2+y-xy-y^2-2xy+2x^2y+2xy^2-\frac7{27}≤0$ $x^2-x-y+xy+y^2+2xy-2x^2y-2xy^2+\frac{7}{27}≥0$ $x^2(1-2y)+x(3y-2y^2-1)+(y^2-y+\frac7{27})≥0$ $2(\frac12-y)\left(x+\frac{y-1}{2}\right)^2+\frac{1}{108} (3y-1)^2(6y+1)≥0$  Of course, I'm not sure the solution is correct.  Can you verify the solution? Thank you.
104836
I have to prove the following inequality:     $$\forall n,m \in \mathbb{N},i\in\mathbb{N_0}:n&lt;m \implies \frac1{n^i}\binom{n}i \leq\frac1{m^i}\binom{m}i $$   I used this fact to prove an other inequality, without proving this one first, but now I am stuck. Is induction the way to do this?
102344
How to prove the following inequality : $$ \sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3 $$ with $a&gt;0,\  b&gt;0$ and $c&gt;0$.
41446
A proof of the inequality using properties of $\pi$ and $e$, for example, is what I'm looking for. Not calculator approximations showing the inequality holds.
114720
I got stuck on this inequality for a day. If $p$ is positive integer, then the problem becomes too easy, but I can't find how we deal with the general case when p can be any positive number. Can someone give me a hint? I really appreciate     Prove that $|x^p - y^p| \le p|x-y|(x^{p-1} + y^{p-1})$ provided that $1 \le p \lt \infty$ and $x, y \ge 0$
8358
In the pdf which you can download  I found the following inequality which I can't solve it.     Exercise 2.1.11 Let $a,b,c  \gt 0$. Prove that    $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}.$$   Thanks :)
91951
Any hints on how I can prove the inequality:     $$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$$   Thank you.
92407
Let $(X,d)$ be a metric space.  Define $$d'((x,y),(z,w))=max\{d(x,z),d(y,w)\}.$$   I'm trying to prove the triangle inequality for this, but really don't have a clue how.  Any tips or suggestions would be greatly appreciated!
99394
$A$$\rightarrow$$(B$ $\vee$ $C$ ) , $B$ $\rightarrow$ $C$ $\vDash$ $A$ $\rightarrow$ $D$  I think it's wrong but I have no idea how to prove.
79745
The inequality is homogeneous, but I am not sure if that helps. Also, I was thinking about considering a concave function and apply Jensen or Karamata on LHS.
2180
Show that $|x-y|\leq |x|+|y|$.   I know this probably applies the triangle inequality in some way, but I can't figure it out.
98279
I have derived the inequality if $k&gt;1$,  ${n(n-1)⋯(n-k+1)\over k!} ({1\over n})^k&lt;{(n+1)n⋯(n-k+2)\over k!} ({1\over n+1})^k$  But, my problem is how to use this inequality to prove that if $n\geq1$, $(1+{1\over n})^n&lt;(1+{1\over n+1})^{n+1}$
106114
pp  and ,   illustrates Triangle Inequality. But without referring to Triangle Inequality at all, how to picture  Reverse Triangle Inequality for all $\mathbf{a,b} \in C^n$? I don't want proofs or rigour.
38566
Prove that $\forall x,y&gt;0, x^x+y^y \geq x^y + y^x$   A friend of mine told me none of the teachers in my school have succeeded in proving this seemingly simple inequality (it was asked at an oral exam last year). I tried it myself, but I made no significant progress.
81975
Let $a,b \in [0,1], a+b=1$ and let $\{u,v\}\subset\mathbb{R}$.   How can we prove that $$(au+bv)^2\leq au^2+bv^2?$$   Opening the brackets give me in the left side $(au+v-av)^2$, but I'm not think that's the right way to do that.  I don't find the triangle inequality very helpfull, too.  Thanks.