Ngoc Hung

Bài 3. Từ giả thiết $a+b+c=3abc\Rightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=3$

Đặt $\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\Rightarrow xy+yz+zx=3$ và x, y, z > 0.

Khi đó $P=\frac{x}{x+1+3yz}+\frac{y}{y+1+3zx}+\frac{z}{z+1+3xy}$

$P=\frac{x^{2}}{x^{2}+x+3xyz}+\frac{y^{2}}{y^{2}+y+3xyz}+\frac{z^{2}}{z^{2}+z+3xyz}$

$\geq \frac{\left ( x+y+z \right )^{2}}{9xyz+x^{2}+y^{2}+z^{2}+x+y+z}$

Áp dụng Cauchy, ta có $\left ( x+y+z \right )\left ( xy+yz+zx \right )\geq \sqrt[3]{xyz}.\sqrt[3]{x^{2}y^{2}z^{2}}=9xyz$

$\Rightarrow 9xyz\leq 3\left ( x+y+z \right )\Leftrightarrow 3xyz\leq x+y+z$

Lại có $xy+yz+zx\geq 3\sqrt[3]{x^{2}y^{2}z^{2}}\Rightarrow xyz\leq 1$

Suy ra $9xyz=3xyz+6xyz\leq x+y+z+2\left ( xy+yz+zx \right )$

$\Rightarrow 9xyz+x^{2}+y^{2}+z^{2}+x+y+z\leq \left ( x+y+z \right )^{2}+2\left ( x+y+z \right )$

Lại có $\left ( x+y+z \right )^{2}\geq 3\left ( xy+yz+zx \right )=9\Rightarrow x+y+z\geq 3$

Vậy $P\geq 1-\frac{2}{3+2}=\frac{3}{5}$

Ta có $a=\frac{5-4b}{3}=1-b+\frac{2-b}{3}$ là số nguyên. Đặt $\frac{2-b}{3}=t\Rightarrow b=2-3t$

Suy ra $a=4t-1$. Khi đó $P=7\left | 4t-1 \right |-2\left | 2-3t \right |$

Xét $t\leq 0\Rightarrow P=-22t+3\geq 3$

Xét $t\geq 1\Rightarrow P=22t-3\geq 19$

Ta có $Q=\left ( x+1 \right )\left ( \frac{1}{\sqrt{x^{2}+3}}+\frac{1}{\sqrt{3x^{2}+1}} \right )$

$=\frac{\left ( x+1 \right )\left ( \sqrt{x^{2}+3}+\sqrt{1+3x^{2}} \right )}{\sqrt{x^{2}+3}.\sqrt{1+3x^{2}}}\leq \frac{\left ( x+1 \right )\left ( 2\sqrt{2}\sqrt{x^{2}+1} \right )}{\sqrt{x^{2}+3}.\sqrt{1+3x^{2}}}$

Xét $x+1<0\Rightarrow Q<0$. Xét $x+1\geq 0\Leftrightarrow x\geq -1$

Khi đó $\frac{\left ( x+1 \right ).\sqrt{x^{2}+1}}{\sqrt{x^{2}+3}.\sqrt{3x^{2}+1}}\leq \frac{1}{\sqrt{2}}\Leftrightarrow 2\left ( x^{2}+1 \right )\left ( x+1 \right )^{2}\leq \left ( x^{2}+3 \right )\left ( 3x^{2}+1 \right )\Leftrightarrow \left ( x-1 \right )^{4}\geq 0$

Do đó $Q\leq 2\leftrightarrow x=1$

Ta có $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6$

$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$

$\frac{1}{a^{2}}+1\geq \frac{2}{a};\frac{1}{b^{2}}+1\geq \frac{2}{b};\frac{1}{c^{2}}+1\geq \frac{2}{c}$

Suy ra $$3\left ( \frac{1}{a^{2}} +\frac{1}{b^{2}}+\frac{1}{c^{2}}\right )+3\geq 2\left ( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )=12\Leftrightarrow \frac{1}{a^{2}} +\frac{1}{b^{2}}+\frac{1}{c^{2}}\geqslant 3$$

Cho $a, b, c$ thỏa mãn $a^2 + b^2 + c^2=1$  chứng minh rằng $abc + 2(1+a+b+c+ab+ac+bc) \geq 0 $

Từ $a^{2}+b^{2}+c^{2}=1\Rightarrow -1\leq a;b;c\leq 1$

Do đó $\left ( 1+a \right )\left ( 1+b \right )\left ( 1+c \right )\geq 0$

$\Rightarrow abc+ab+bc+ca+a+b+c+1\geq 0$

Mặt khác $1+a+b+c+ab+bc+ca=\frac{1}{2}\left ( a+b+c+1 \right )^{2}\geq 0$

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