Let $x,y,z \in \Bbb R^+$ such that $x+y+z=3$. Prove the inequality   $\sqrt x+\sqrt y+\sqrt z \ge xy+yz+zx$     I tried to prove that $\sqrt x+\sqrt y+\sqrt z-(xy+yz+zx)\ge 0$ I squared the equality, put the value of $xy+yz+zx$ (in terms of $x^2+y^2+z^2$). Now I tried to prove that the above expression but failed.    Thanks for hints or solutions.
43018
Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.  Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$    Here's my idea:  $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  And I'm stuck here.  I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$  In the first case using Cauchy-Schwarz Inequality I prove that:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $3(a^2 + b^2 + c^2) \ge (a+b+c)^2$  Now I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$   $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$   I need I don't know how to continue.  In the second case I tried proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and  $a^2 + b^2 + c^2 \ge a+b+c$  Using Cauchy-Schwarz Inequality I proved:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$  $a^2 + b^2 + c^2 \ge a+b+c$  But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$  So please help me with this problem.  P.S   My initial idea, which is proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  maybe isn't the right way to prove this inequality.
64023
let $x,y,z,w&gt;0$,show that $$(x+y)^3+(y+z)^3+(z+w)^3+(w+x)^3\ge 8(x^2y+y^2z+z^2w+w^2x)$$  it seem use AM-GM inequality to solve it,But I can't it,Thanks
48128
Verify that  $\sqrt{2}\left\| z \right\| \ge \left|\Re(z)\right| + \left|\Im(z)\right|.$  I started off noting that $z=x+iy$ and that $Re(z)=x$ and $Im(z)=y$  Then I know that I have to square both sides, giving me   $$2(x^2 +y^2) \ge x^2+2|xy|+y^2$$  I'm not sure what to do after this.
13673
$x,y,z &gt;0$, prove   $$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$   Note: Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.   Update 1: In this forum, somebody said  that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.   Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by  in  may help.    Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.  Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".
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.
34863
Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$.   My Attempt:  We will calculate for $x,y,z&gt;0$. Then, using the AM-GM Inequality, we have  $$ \begin{cases} x^2+y^2\geq 2xy\\ y^2+z^2\geq 2yz\\ z^2+x^2\geq 2zx\\ \end{cases} $$  So $x^2+y^2+z^2\geq xy+yz+zx$. How can I solve for $(x,y,z)$ after this?
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.
102884
I need to prove or disprove the following inequality: $$ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $$ For $x,y,z \in \mathbb R^+$. I found no counter examples, so I think it should be true. I tried Cauchy-Schwarz, but I didn't get anything useful. Is it possible to prove this inequality without using brute force methods like Bunching and Schur?  This inequality was in the Iran MO in 1996.
8478
let $x,y,z&gt;0$ and such $xy+yz+zx=3$,show that $$\dfrac{x}{x^3+y^2+1}+\dfrac{y}{y^3+z^2+1}+\dfrac{z}{z^3+x^2+1}\le 1$$  To prove this inequality,I want use following Cauchy-Schwarz inequality $$(x^3+y^2+1)(\frac{1}{x}+1+z^2)\ge (x+y+z)^2$$ $$\dfrac{x}{x^3+y^2+1}\le \dfrac{1+x+xz^2}{(x+y+z)^2}$$ we have $$\sum\dfrac{x}{x^3+y^2+1}\le\dfrac{3+x+y+z+xz^2+yx^2+zy^2}{(x+y+z)^2}$$ it suffices to prove that  $$x+y+z+xz^2+yx^2+zy^2\le x^2+y^2+z^2+3$$
106384
An inequality proposed at Zhautykov Olympiad 2008.     Let be $a,b,c &gt;0$ with $abc=1$. Prove that:    $$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$   Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.   Our inequality becomes: $$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$  Now we use that: $z^2+x^2 \geq 2zx.$ $$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \sum_{\mathrm{cyc}}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$  Now applying Cauchy-Schwarz we obtain the desired result.  What I wrote can be found on this link: . But now, I don't know how to apply Cauchy-Schwarz.   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.
97881
prove that $x^2 + y^2 + z^2 \ge xy + xz + yz$  I tried proving it the way I prove $x^2 + y^2 \ge 2xy$. But it just doesn't seem to work. I tried using AM-GM. Should I be using GM-HM or AM-HM? I really don't know what the problem is and I'd appreciate any help and a full proof explanation. Thanks.
68929
Let $x,y,z&gt;0$: Prove that:   $\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}$  I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!
61045
This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).  If $x, y, z$ are positive real numbers, prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$  My best idea was to expand this and simplify. Although that doesn't look very feasible. Another idea is to see that $x^2+y^2+xy \geq x^2+y^2$. Then we just have to show that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+y^2)(x^2+z^2)(y^2+z^2)$ if that is even true.
87993
Let $a$, $b$ and $c$ be non-negative numbers such that $a^3+b^3+c^3=3$. Prove that:   $$a^4b+b^4c+c^4a\leq3$$   This inequality similar to the following.     Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that:   $$a^3b+b^3c+c^3a\leq3,$$   which follows from the following identity.   $$(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)=\frac{1}{2}\sum_{cyc}(a^2-b^2-ab-ac+2bc)^2.$$   I tried Rearrangement.  Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.  Hence, $$a^4b+b^4c+c^4a=a^3\cdot ab+b^3\cdot bc+c^3\cdot ca\leq x^3\cdot xy+y^3\cdot xz+z^3\cdot yz=$$ $$=y(x^4+y^2xz+z^4)$$ and I don't see what is the rest.  Thank you!
58414
I want to prove the following inequality :      $$\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab,  $$   $a,b,c,d \in \mathbb{R}_{+} .$    In my book, at the answers chapter the author uses AM $\geq$ GM, but I haven't any idea how I can use that.   Thanks :)
71323
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive   $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$  $$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$  Is the proof valid?