For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$ I tried to pls around trying to reorganize to get AM-GM but i couldn't  Thanks for the help in advance.
120544
Prove that if a,b.c are distinct positive integers that $$a^4+b^4+c^4&gt;abc(a+b+c)$$ My attempt:  I used the inequality A.M>G.M to get two inequalities First inequality $$\frac{a^4+b^4+c^4}{3} &gt; \sqrt[3]{a^4b^4c^4}$$  or $$\frac{a^4+b^4+c^4}{3} &gt; abc \sqrt[3]{abc}$$ or new -- first inequality $$\frac{a^4+b^4+c^4}{3abc} &gt; \sqrt[3]{abc}$$  second inequality: $$\frac{a+b+c}{3} &gt; \sqrt[3]{abc}$$ I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??
60943
If $a,b,c &gt; 0$ are such that $abc=1$, then   $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c. $$   I would be pleased if you give me a hint. Thanks in advance.
87868
$a,b,c,d&gt;0$ with $a+b+c+d=4$ then $\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2$   How can I apply AM-GM here, is there a generalization, say we have $n$ variables summing up to $n$ then,  $\sum\limits_{cyc}a_1\cdot a_2\dots a_k\le\frac1{n^{k-1}}\left(\sum\limits_{cyc}a\right)^k$  because the following inequality should be also true  $\sum\limits_{cyc}abc\le\frac1{16}\left(\sum\limits_{cyc}a\right)^3$
55741
How one can prove the following.  Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:  $(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$  WLOG one can assume that $0\le a\le b\le c$.  It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le a\le b\le c$.
21307
If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.
99756
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that    $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$     My work: $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}$$ Now, we have $a^2+\dfrac{1}{a^2}\ge 2$ and $b^2+\dfrac{1}{b^2}\ge 2$. Here, I am stuck, I cannot use the information provided, $a+b=1$ to any use. Please help!
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.
114307
$a, b, c$ are positive real numbers.   Prove that $$8abc\le (a+b)(b+c)(c+a)$$  I haven't really gotten anywhere, just tried to open up $(a+b)(b+c)(c+a)$ and got null.
95127
I just started doing AM-GM inequalities for the first time about two hours ago. In those two hours, I have completed exactly two problems. I am stuck on this third one! Here is the problem:  If $a, b, c \gt 0$ prove that $$ a^3 +b^3 +c^3 \ge a^2b +b^2c+c^2a.$$  I am going crazy over this! A hint or proof would be much appreciated. Also any general advice for proving AM-GM inequalities would bring me happiness to my heart. Thank you!
55163
Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d = 4$. Prove that $$\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a .$$   How can I approach this using only the ? Are there any other methods that do not involve the AM-GM inequality?
76815
Let $a,b,c$ be positive real numbers satisfying $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16,$$ then $a+b+c=3$.  I thought about the application of the AM-GM-HM inequality, but in vain. I also thought about splitting 16 into factors and comparing but went nowhere. Any ideas. Thanks beforehand.
54416
Disclaimer: The statement may be false, but for now I'm operating under the assumption it's true and trying to prove it.   My workings:  I got a common denominator and expressed it as:  $$\frac{ab^2 + bc^2 + ca^2}{abc} \ge 3$$ $$ab^2 + bc^2 + ca^2 \ge 3abc$$  This is where I'm stuck. I feel as if I keep encountering this same result in various different problems and always am stumped on how to show it is true.   I tried to look at similar problems and it seems AM-GM inequality is the usual line of attack, so using that I reasoned, $$a^2 + b^2 + c^2 \ge 3abc$$, and now I need to establish $$ca^2 + ab^2 + bc^2 \ge a^2 + b^2 + c^2$$  However, this doesn't seem like it'd be true for general positive $a,b,c$.
109429
For positive real numbers $a,b,c$ prove that $$(a+b+c)(ab+bc+ca)(a^3+b^3+c^3)\le (a^2+b^2+c^2)^3$$ My try : Rewrite this as $$\frac{a^2+b^2+c^2}{ab+bc+ca}-1\ge \frac{(a+b+c)(a^3+b^3+c^3)}{a^2+b^2+c^2)^2}-1$$ Then I am stumped, I was trying to use SOS but I think an easier proof is possible. My teacher give this to me. He combined to well known inequalities $a^2+b^2+c^2\ge ab+bc+ca$ and $(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)^2$ and changed the sign. This is easy by  but my teacher wouldn't accept it. Can some one help me? Thanks in advance.
76501
For $a,b,c&gt;0$, prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}.$$ (Source : Balkan Mathematical Olympiad 2005)  My work: From the given inequality, we can have, $$\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}\ge \frac{4(a-b)^2}{a+b+c}$$ This can be re-written as, $$(a+b+c)\left(\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}\right)\ge 4(a-b)^2$$ Now, what am I supposed to do? I am doubtful whether the question is correct or not. Please help. EDIT: After checking quite a few cases I conclude that the inequality is actually working. So there might not be any mistakes at all.
71836
Let $a, b, c, d &gt; 0$. Prove that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$.   I'm supposed to prove this by AM-GM, but I can't see how. Any help would be appreciated.
21037
Problem. Prove that if $ a,b,c &gt; 0 $, then $ [(1 + a) (1 + b) (1 + c)]^{7} &gt; 7^{7} (a^{4} b^{4} c^{4}) $.   I don’t know how to solve this problem... What I can think of is to just simplify this inequality: $$ \left[ \frac{(1 + a) (1 + b) (1 + c)}{7} \right]^{7} &gt; a^{4} b^{4} c^{4}. $$ How can I proceed with solving this problem?    Note: This is a question of sequence and series, specifically AM-GM-HM inequality...
40499
If $a, b, c$ belong to positive real numbers then prove that  $$\frac{2}{b+c} +\frac{2}{c+a}+\frac{2}{a+b} \le \frac 1a +\frac 1b +\frac 1c.$$  I applied AM-GM inequalities on both sides but I cant even figure out what to do on left side.