Let $a,b,c&gt;0$; $a^2+b^2+c^2=3$. Prove that: $$\sum\limits_{cyc} \sqrt{a^2+3b^2}\geq \sqrt{12(a+b+c)}$$ I have got stuck at this problem. Any soluion is welcome.
122966
Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b+c)^2(a^2+b^2+c^2)=27$. Prove that: $$\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$$  A big problem here around $(a,b,c)=(1.6185...,0.71686...,0.4926...)$.  In this case we get $\sum\limits_{cyc}\sqrt{a^2+3b^2}-6=0.000563...$.  My trying.   Let $a^2+3b^2=4x^2$, $b^2+3c^2=4y^2$ and $c^2+3a^2=4z^2$, where $x$, $y$ and $z$ are non-negatives.  Hence, we need to prove that $$\sum\limits_{cyc}\sqrt{x^2-3y^2+9z^2}\leq\frac{\sqrt7(x+y+z)^2}{\sqrt{3(x^2+y^2+z^2)}}$$  Let $k$ and $m$ be non-negatives, for which  $x-ky+mz&gt;0$, $y-kz+mx&gt;0$, $z-kx+my&gt;0$ and $1-k+m&gt;0$.  By C-S $\left(\sum\limits_{cyc}\sqrt{x^2-3y^2+9z^2}\right)^2\leq(1-k+m)(x+y+z)\sum\limits_{cyc}\frac{x^2-3y^2+9z^2}{x-ky+mz}$.  Thus, it remains to prove that $$(1-k+m)\sum\limits_{cyc}\frac{x^2-3y^2+9z^2}{x-ky+mz}\leq\frac{7(x+y+z)^3}{3(x^2+y^2+z^2)}$$  It's a sixth degree, but I didn't find a values of $k$ and $m$, such that the last inequality will be true.  By this way we can prove that $\sum\limits_{cyc}\sqrt{a^2+2b^2}\geq3\sqrt3$ is true, but it's not comforting.  Also I tried to use Holder, but without success.  Thank you!
56179
Problem :  Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.  My approach : Since : $(a^3+b^3)=(a+b)^3-3ab(a+b)$ $\Rightarrow z^3+\frac{1}{z^3}=(z+\frac{1}{z})^3-3(z+\frac{1}{z})$ Now I don't know further about this problem whether or not this will help here. Please guide to solve this problem will be of great help, thanks.
92621
Let $a,b,c$ are real number such that $a+b+c=1$. Prove that: $$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \leq \frac{9}{10}.$$
81876
$a,b,c$ are real numbers $&gt;0$. If $a+b+c=1$, show that     $$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\geq\frac{100}{3}$$
91562
$a,b,c$ are positive real numbers such that $a+b+c=1$. Prove that: $$\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2} \geqslant \dfrac{1}{2} $$ I have tried with Cauchy-Schwarz inequality in Engel form...
84254
If $a,b,c$ real numbers such that $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $$,  prove that $ab + bc + ca \le 3$.     Attempt:  Assume that $a,b,c &gt; 0$.   $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1  $$ $$ \implies  (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1) + (a^{2} + b^{2} + 1)(a^{2}+c^{2}+1) +  (a^{2} + c^{2} + 1)(b^{2}+c^{2}+1) \ge (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1)$$  $$ \implies  ((ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 b^{2} + c^{2} + b^{4} + 1) + ((ab)^{2} + (ac)^{2} + (bc)^{2} + b^{2} + 2 a^{2} + c^{2} + a^{4} + 1) + ((ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 c^{2} + b^{2} + c^{4} + 1) \ge (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1)$$  $$ \implies  3((ab)^{2} + (ac)^{2} + (bc)^{2}) + 4(a^{2} + b^{2} + c^{2}) + (a^{4}+b^{4}+c^{4}) + 3  \ge (a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1)$$  Next, notice  $$(a^{2} + b^{2} + 1)(b^{2}+c^{2}+1)(c^{2} + a^{2} + 1) = ((ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 b^{2} + c^{2} + b^{4} + 1)(c^{2} + a^{2} + 1) $$ $$ = (abc)^{2} + a^{2}c^{4} + b^{2}c^{4} + (ac)^{2} + 2(bc)^{2} + c^{4} + b^{4}c^{2} + c^{2} + a^{4} b^{2} + a^{4}c^{2} + (abc)^{2} + a^{4}  + 2(ab)^{2} + (ac)^{2} + b^{4}a^{2} + a^{2} +  (ab)^{2} + (ac)^{2} + (bc)^{2} + a^{2} + 2 b^{2} + c^{2} + b^{4} + 1 $$ $$ = 2(abc)^{2} + 3(ac)^{2} + 3(bc)^{2} + 3(ab)^{2} + c^{4}a^{2} + c^{4} b^{2} + b^{4}a^{2} + b^{4} c^{2} + a^{4}b^{2} + a^{4} c^{2} + a^{4} + b^{4} + c^{4} + 2(a^{2} + b^{2} + c^{2}) + 1  $$  So we have  $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2}+c^{2}+1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $$ $$ \implies  2(a^{2} + b^{2} + c^{2}) +  3  \ge 2(abc)^{2}  + c^{4}a^{2} + c^{4} b^{2} + b^{4}a^{2} + b^{4} c^{2} + a^{4}b^{2} + a^{4} c^{2} + 1 $$
109760
$a, b,c $ are positive real numbers such that $a+b+c = 3$, prove that :$a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6$  Any ideas ?
114077
How can we show that the assumption $a,b,c&gt;0$ and $a+b+c=1$ implies   $$\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}~?$$
108090
Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$,  $b^2+c^2=9$  and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$?  Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find $a,b,c$
25531
$a,b,c$ are positive reals with $abc = 1$. Prove that    $$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$   I try to use AM $\ge$ HM.  $$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{a^3(b+c)+b^3(a+c)+c^3(a+b)}$$ Then how I proceed.
41115
Let $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=168$. Find $a,b,c$   Suppose $a, b, c$ are roots of $P(x)$.   $$P(x) = k(x - a)(x - b)(x - c)$$  But then I get $(k = 1)$  $$P(x) = x^3 - 12x^2 + x(ab + ac + bc) - abc$$  Cant go further...
48670
$a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$    My work: I tried using Cauchy-Schwarz inequality to find that, $(a^2+b^2+c^2)(1^2+1^2+1^2)\ge (a+b+c)^2$ $(a^2+b^2+c^2)\ge \dfrac13(a+b+c)^2$ $\sqrt{(a^2+b^2+c^2)}\ge \dfrac{1}{\sqrt{3}}(a+b+c)\ge \dfrac{1}{\sqrt{3}}abc$ which is not what I need and neither I can use it to prove the required inequality. Please help.
22071
I did substitute $a=x+y, b=x+z, c=y+z$ and I arrived at $\sqrt{2x} + \sqrt{2y} + \sqrt{2z} \le \sqrt{x+y} + \sqrt{x+z} + \sqrt{y+z}$. However, after this, I tried various methods like AM-GM and Cauchy-Schwarz inequality for hours and I still can't prove it. Can someone help please? Thanks.
58658
if $a,b,c$ are non zero positive reals prove $$\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$$ I have used A.M  G.M  inequality as follows:  $$a^3+b^3+c^3 \ge 3abc$$ adding $abc$ both sides we get  $$a^3+b^3+abc \ge 4abc-c^3=c(4ab-c^2)$$ so  $$\frac{1}{a^3+b^3+abc} \le \frac{1}{c(4ab-c^2)}$$ But since $a,b,c$ are positive reals  $$4abc-c^3 &lt; 4abc$$ so  $$\frac{1}{4abc-c^3} \gt \frac{1}{4abc}$$  but i am unable to proceed here
50944
Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$  I tried it to do using $AM \ge GM$ but don't know how to proceed.  Please help.
63677
let $a,b,c&gt;0$. Show that   $$a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$$   By the $$(1+a)(1+b)(1+c)=1+a+b+c+ab+bc+ac+abc$$ it is suffient to show  $$a^2+b^2+c^2+abc+2\ge a+b+c+ab+bc+ac$$ How?
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.