Mladenov wrote: Prove that if $a,b,c \ge 1$ and $a+b+c=9$, then $\sqrt{ab+bc+ca} \le \sqrt{a} +\sqrt{b} + \sqrt{c}$ See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=105823

2012 Chern Cup National High School Mathematical Olympiad, Day 2, Problem 1 (China Tianjin)

Proof by Van Tien Nguyen, Hoang Van Thu high school 2014-2017, Viet Nam Inequality is equivalent to: $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2 \ge ab+bc+ca$$or $$a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c})^2 \ge a^2+b^2+c^2+2(ab+bc+ca)=(a+b+c)^2=81$$ Use AM-GM we have: $$2(\sqrt{a}+\sqrt{b}+\sqrt{c})^2+54 \ge 12\sqrt{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})$$so we only need prove: $$a^2+b^2+c^2+12\sqrt{3}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \ge 135 (1)$$ Let $$x=\sqrt{3a} \ge \sqrt{3}, y=\sqrt{3b} \ge \sqrt{3}, z=\sqrt{3c} \ge \sqrt{3}$$we have $$x^2+y^2+z^2=27$$. Inequality (1) equivalent to: $$\sum \frac{x^4}{9}+12 \sum x \ge 135$$ Consider inequality: $$\frac{x^4}{9}+12x \ge 4x^2+9$$ or $$(x-3)^2(x^2+6x-9) \ge 0$$true because $$x^2+6x-9 \ge 3+6\sqrt{3}-9=6\sqrt{3}-6 > 0$$ From that, easy to done.

Let $a=x^2, b=y^2,c=z^2 \ \ \text{and} x\ge y\ge z\ge 1$. Then $x^2+y^2+z^2=9$. We need to prove $(x+y+z)^2\ge (xy)^2+(yz)^2+(zx)^2 $ $ \Leftarrow (x^2+y^2+z^2)(x+y+z)^2\ge 9((xy)^2+(yz)^2+(zx)^2)$ $\iff (x - y)^2 (2 x^2 + 9 x y + 3 x z + 2 y^2 + 3 y z - 7 z^2)+ $ $(z-x)(z-y)(2z^2+6zx+6zy-x^2-y^2)\ge 0$ $\Leftarrow 2z^2+6zx+6zy-x^2-y^2\ge 0$ $\Leftarrow 3z^2+6zx+6zy-9\ge 0$ which is obviously true due to $x\ge y\ge z\ge 1$

sqing wrote: 2012 Chern Cup National High School Mathematical Olympiad, Day 2, Problem 1 (China Tianjin) Let $a,b,c> 1$ and $a+b+c=9$. Prove that: $$ab+bc+ca\leq \sqrt{a}+\sqrt{b}+\sqrt{c}$$Shiing-Shen Chern Cup Mathematical Olympiads 2012，Q6 Prove that if $a,b,c \ge 1$ and $a+b+c=9$, then $$ab+bc+ca \le 3(\sqrt{ab} +\sqrt{bc} + \sqrt{ca})\leq ( \sqrt{a}+\sqrt{b}+\sqrt{c})^2.$$

sqing wrote: sqing wrote: 2012 Chern Cup National High School Mathematical Olympiad, Day 2, Problem 1 (China Tianjin) Let $a,b,c> 1$ and $a+b+c=9$. Prove that: $$ab+bc+ca\leq \sqrt{a}+\sqrt{b}+\sqrt{c}$$Shiing-Shen Chern Cup Mathematical Olympiads 2012，Q6 Prove that if $a,b,c \ge 1$ and $a+b+c=9$, then $$ab+bc+ca \le 3(\sqrt{ab} +\sqrt{bc} + \sqrt{ca})\leq ( \sqrt{a}+\sqrt{b}+\sqrt{c})^2.$$ Macedonian Olympiad If $a,b,c>1$ and $a+b+c=9$; prove that$$\sqrt{ab+ac+bc}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}\leq \sqrt{{a^{2}+b^{2}+c^{2}}}$$ Let $a,b,c\geq\sqrt{\frac{13-4\sqrt{10}}{3}}$ and $a^2+b^2+c^2=9.$ Prove that$$a^2b^2+b^2c^2+c^2a^2\leq (a+b+c)^2.$$p/174064635455

Find the smallest constant $m > 0$ such that for any $a, b, c \geq m, a+b+c=9$, we always have $$\sqrt{ab+bc+ca} \leq \sqrt{a}+\sqrt{b}+\sqrt{c}$$

Notice that $\sum_{cyc}ab\le\frac{\left(\sum_{cyc}a\right)^2}{3}=27$, thus the inequality is equivalent to $\sqrt{27}\le\sum_{cyc}\sqrt{a}\Longleftrightarrow 3\sqrt{3}\le\sum_{cyc}\sqrt{a}$ Furthermore let $a=x^2, b=y^2\text{ and }c=z^2$, therefore the inequality and the condition become $x+y+z\ge 3\sqrt{3}\text{ and }x^2+y^2+z^2=9$ Now let $f(x,y,z)=x+y+z, g(x,y,z)=x^2+y^2+z^2-9\text{ and let }L=f(x,y,z)-\lambda g(x,y,z)$ $$\frac{\partial L}{\partial x}=1-2\lambda x=0\Longrightarrow 2\lambda x=1\Longleftrightarrow x=\frac{1}{2\lambda}$$$$\frac{\partial L}{\partial y}=1-2\lambda y=0\Longrightarrow 2\lambda y=1\Longleftrightarrow y=\frac{1}{2\lambda}$$$$\frac{\partial L}{\partial z}=1-2\lambda z=0\Longrightarrow 2\lambda z=1\Longleftrightarrow z=\frac{1}{2\lambda}$$$$\text{ therefore }x=y=z$$Furthermore from $g(x,x,x)$ we obtain $3x^2-9=0\Longrightarrow x^2=3\Longrightarrow x=y=z=\sqrt{3}$ Thus minimum is obtained in the assertion $f(\sqrt{3},\sqrt{3},\sqrt{3})=\sqrt{3}+\sqrt{3}+\sqrt{3}=3\sqrt{3}$ $\blacksquare$.

Mladenov wrote: Prove that if $a,b,c \ge 1$ and $a+b+c=9$, then $$\sqrt{ab+bc+ca} \le \sqrt{a} +\sqrt{b} + \sqrt{c}$$

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Let $a,b,c$ be nonnegative real numbers such that $a^2+b^2+c^2=27 $. Prove that $$\sqrt{ab+bc+ca} \le \sqrt{a} +\sqrt{b} + \sqrt{c}$$

arqady wrote: Mladenov wrote: Prove that if $a,b,c \ge 1$ and $a+b+c=9$, then $\sqrt{ab+bc+ca} \le \sqrt{a} +\sqrt{b} + \sqrt{c}$ See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=105823 Thank you

Let $(x,y,z)=(\sqrt{a},\sqrt{b},\sqrt{c})$, $9x^2\ge x^2+y^2+z^2\ge x^2+\frac{1}{2}(y+z)^2$, $4x\ge y+z$. $(x+y+z)^2(x^2+y^2+z^2)-9(x^2y^2+x^2z^2+y^2z^2)=\sum{(xy+(4x-y-z)(4y-x-z))(x-y)^2}+5(x^2+y^2+z^2-xy-xz-yz)^2\ge0$.