For the non-negative real numbers $a, b, c$ prove that $$(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$$   What I did is applying Holder's inequality in LHS:$$(a^2+(\sqrt{2})^2)(b^2+(\sqrt{2})^2)(c^2+(\sqrt{2})^2) \geq (abc + 2\sqrt{2})^2$$   Then it suffices to prove that $$(abc+2\sqrt2)^2 \geq 3(a+b+c)^2 \\ \Rightarrow abc+2\sqrt2 \geq \sqrt3(a+b+c)$$   But I don't know how to proceed. I also think I applied Holder's Inequality incorrectly.
17451
Let $a$, $b$, $c$ be nonnegative real numbers.   Prove $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$.
51490
I stumbled on the following inequality: For all $n\geq 1,$ $$2\sqrt{n+1}-2\sqrt{n}&lt;\frac{1}{\sqrt{n}}&lt;2\sqrt{n}-2\sqrt{n-1}.$$ However I cannot find the proof of this anywhere. Any ideas how to proceed?  Edit: I posted a follow-up question about generalizations of this inequality here:
5974
Let $a,b,$ and $c$ be nonnegative real numbers such that $a^2+b^2+c^2+abc = 4$. Prove that $$0 \leq ab + ac + bc - abc \leq 2.$$   I tried using rearrangement to get $a^2+b^2+c^2+abc = 4 \geq ab+bc+ac+abc$. Then I just need to show that $0\leq ab + ac + bc - abc$ and $4-2abc \leq 2$. I am not sure if this method will work, though.
50245
Suppose that $a,b,c$ are the lengths of the sides of a triangle. Prove that $$a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$$   Attempt  It's hard to do much simplification here so we are going to have to use that $a &gt; b+c, b&gt;a+b, c&gt;a+c.$ Using that fact we obtain that $$b^2(c-b)+c^2(b-c) &gt; a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c).$$ Then should I use AM-GM to finish it or what inequality?
94331
Problem: Hi, I'm trying to solve the following proof:  If $a &gt; b \geq 0$, Prove that $a^2 &gt; b^2$.  But I'm a bit new with proofs and I don't manage to write a proper one. I would be very grateful if someone could explain me which are the steps required to demonstrate the proof. Thank you.
114584
Prove inequality with $a,b,c &gt;0$ $$\frac{a^4}{a^3+b^3} +  \frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3} \ge \frac{a+b+c}{2}$$   I tried the inequality: $\sum \frac {a^4+b^4}{a^3+b^3} \ge \sum \frac{a+b}2=a+b+c$, but seem like it doesn't help
45742
For positive integers Numbers  $a, b, c $ prove that  $$\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq  2$$ This inequality above take long time to prove it and still couldn't complete it.  How I can prove this inequality?  Any hint will help. Thanks
57583
Let $a,b,c&gt; 0$ be three real numbers such that $a+b+c=1$. I want to prove that   $$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\le\frac{1}{5abc}.$$  My attempt: Using AM-GM on each denominator gives (here, LHS denotes the left-hand side) $$LHS\le\frac{1}{3a^{\frac12}bc}+\frac{1}{3b^{\frac12}ac}+\frac{1}{3c^{\frac12}ab}.$$  However, I think that my attempt doesn't work because the last expression can get larger than $\frac{1}{5abc}$. In fact, if we multiply with $3abc$ then the original inequality is: $$\sqrt a+\sqrt b+\sqrt c\le\frac35$$  However, as $a,b,c&lt;1$, $$\sqrt a +\sqrt b + \sqrt c&gt; a+b+c=1&gt;\frac35.$$ So my upper bound is always larger than the given one.
119945
I'm about to prove that for any $a,b&gt;0$ and $n\in\mathbb{N},$ the inequality: $\frac{a^n+b^n}{2}\geq\left(\frac{a+b}{2}\right)^n$ holds.   By induction I get: $$\left(\frac{a+b}{2}\right)\cdot\frac{a^n+b^n}{2}$$ $$=\frac{1}{2}\cdot\frac{a^{n+1}+b^{n+1}+ab^n+ba^n}{2}$$ $$=\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right).$$  Now I have to prove that $$\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right)$$ $$\leq\frac{a^{n+1}+b^{n+1}}{2}$$ $$\Leftrightarrow a^{n+1}+b^{n+1}\geq ab^n +ba^n.$$ But I don't know how to prove that one, is it possible to be proved by induction ? Thank you.
5044
Let $a$, $b$, and $c$ be positive real numbers such that $a+b+c = 1$. Then how to establish the following inequality? $$ (1-a)(1-b)(1-c) \geq 8abc.$$  My effort:   Since $a+b+c =1$, we can write $$  (1-a)(1-b)(1-c) = 1 - (a+b+c) +(ab+bc+ca) - abc = 1 - 1  +(ab+bc+ca) - abc =  (ab+bc+ca) - abc = abc(a^{-1} + b^{-1} + c^{-1} ) - abc = 3abc (\frac{a^{-1} + b^{-1} + c^{-1}}{3}) - abc.$$ What next?
120230
Show that, if $a$ and $h$ are positive numbers, $h &lt; a^2$, then $$\sqrt{a^2 + h}-a &lt; \frac{h}{2a} &lt; a - \sqrt{a^2 - h}$$  I've been working on this problem for about 2 hours now, but I've made no progress. I'm not looking for an answer, but I just need some help to get me started since we didn't practice inequalities this complex in highschool. Thanks.   All I can tell is that we're supposed to take the square root of an expression at some point since one inequality ( h &lt; a^2 ) becomes two.   Edit : Thank you guys for the replies, but I'd appreciate only hints in the future (like Robert Israel) so that I can learn. Regardless, I found a different way to do it, so it's cool :)
86737
So I have to prove $$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$  I rearranged it   $$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$  My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?  Thanks
81550
Prove that the following inequality holds   $$a^a b^b c^c\ge (abc)^{\frac{a+b+c}{3}}$$   if $a,b,c$ are positive.   I'm not sure how to handle these kinds of powers. Are there any "famous" but not so advanced inequalities that involve this kinds of expressions?  I was trying to do something like ordering them without lost of generality because it's symmetric, but I get nowhere and I just make the thing even more complicated.  I'm asking for solution rather than hints since this is my first time encountering these kinds of problems.
85884
Show that for $x,y,z &gt; 0$ the inequality is true: $\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+x+y+z \geq \frac{(x+y)^2}{y+z}+\frac{(y+z)^2}{z+x}+\frac{(z+x)^2}{x+y}$  I have tried Holder, but i had no luck. Please give me a hint of how to start.
113438
Let $a,b,c,d$ be positive real numbers. Prove that $$\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c} \geq \dfrac{2}{3}.$$   I was thinking of trying $a \geq b \geq c$ since the inequality is cyclic. I am not sure how to use this, though, or if this would help simplify the inequality. It also doesn't look like I can really use AM-GM or Cauchy-Schwarz.
23436
If $a$, $b$ and $c$ are real numbers such that $a^2+b^2+c^2=1$ , then how to prove that  $ \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 $  ( don't apply Schur's inequality )? Thanks.
76070
If $a,b,c$ are the sides of a triangle and $p,q,r$ are positive real numbers then prove that $a^2(p-q)(p-r)+b^2(q-p)(q-r)+c^2(r-p)(r-q)\ge 0$  After modification. I have to prove $(a^2 p^2+b^2 q^2 + c^2 r^2) \ge pr(a^2+c^2-b^2)+qr(b^2+c^2-a^2)+pq(a^2+b^2-c^2)$