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Sửa \(\dfrac{1}{3}\rightarrow3\)
Từ \(a+b+c+ab+bc+ca=6abc\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=6\)
Ta có: \(\dfrac{1}{a^2}+1\ge\dfrac{2}{a};\dfrac{1}{b^2}+1\ge\dfrac{2}{b};\dfrac{1}{c^2}+1\ge\dfrac{2}{c}\)
Và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge\dfrac{2}{ab};\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{2}{bc};\dfrac{1}{c^2}+\dfrac{1}{a^2}\ge\dfrac{2}{ac}\)
Cộng theo vế các BĐT trên ta có:
\(3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\Leftrightarrow3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\right)\ge12\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+1\ge4\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\)
\("="\Leftrightarrow a=b=c=1\)
Áp dụng liên tiếp AM-GM và Cauchy-Schwarz ta có:
\(\begin{align*} \dfrac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}&\ge \dfrac{a^2+ab+1}{\sqrt{a^2+ab+c^2+\left (a^2+b^2 \right )}}\\ &=\dfrac{a^2+ab+1}{\sqrt{a^2+ab+1}}\\ &=\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}\\ &=\dfrac{1}{\sqrt{5}}\sqrt{\left ( \dfrac{9}{4}+\dfrac{3}{4}+1+1 \right )\left [\left ( a+\dfrac{b}{2} \right )^2+\dfrac{3b^2}{4}+a^2+c^2 \right ]}\\ &\ge \dfrac{1}{\sqrt{5}}\left [ \dfrac{3}{2}\left (a+\dfrac{b}{2} \right )+\dfrac{3}{4}b+a+c \right ]\\ &=\dfrac{1}{\sqrt{5}}\left ( \dfrac{5}{2}a+\dfrac{3}{2}b+c \right ) \end{align*}\)
Chứng minh tương tự, cộng lại ta có đpcm.
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
bài này cuốn hút thật, lâu lắm ms thấy . xí bài này nhé nghĩ đã lát quay lại làm
Đặt \(x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}, \Rightarrow x+y+z=2\)
Suy ra \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\)
Ta có \(\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{\left(2-x\right)^2} .\frac{2-x}{8}.\frac{2-x}{8}}=\frac{3x}{4}.\)
\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\)\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^2}+\frac{z^3}{\left(2-z\right)^2}\ge x+y+z-\frac{3}{2}=2-\frac{3}{2}=\frac{1}{2}\)
dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)hay \(a=b=c=\frac{3}{2}\)
\(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\sqrt{\frac{ab+2c^2}{a^2+b^2+ab}}=\frac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(a^2+b^2+ab\right)}}\ge\frac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\ge\frac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)
Tương tự: \(\sqrt{\frac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\) ; \(\sqrt{\frac{ca+2b^2}{1+ac-b^2}}\ge ca+2b^2\)
Cộng vế với vế:
\(VT\ge2\left(a^2+b^2+c^2\right)+ab+bc+ca=2+ab+bc+ca\)
Lời giải:
Ta thấy:
\(\text{VT}=(a+\frac{ca}{a+b})+(b+\frac{ab}{b+c})+(c+\frac{bc}{c+a})\)
\(=\frac{a(a+b+c)}{a+b}+\frac{b(a+b+c)}{b+c}+\frac{c(a+b+c)}{c+a}\)
\(=(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\)
\(\geq (a+b+c).\frac{(a+b+c)^2}{a^2+ab+b^2+bc+c^2+ac}=\frac{(a+b+c)^3}{a^2+b^2+c^2+ab+bc+ac}\) (theo BĐT Cauchy-Schwarz)
Có:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2+2$
$\Rightarrow a+b+c=\sqrt{a^2+b^2+c^2+2}=\sqrt{t+2}$ với $t=a^2+b^2+c^2$
Do đó:
$\text{VT}\geq \frac{\sqrt{(t+2)^3}}{t+1}$ \(=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\)
Áp dụng BĐT AM-GM:
\((t+2)^3=\left(\frac{t+1}{2}+\frac{t+1}{2}+1\right)^3\geq 27.\frac{(t+1)^2}{4}\)
\(\Rightarrow \text{VT}=\sqrt{\frac{(t+2)^3}{(t+1)^2}}\geq \sqrt{\frac{27}{4}}=\frac{3\sqrt{3}}{2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{\sqrt{3}}$
\(M\le\frac{1}{4}\Sigma\frac{\left(a+b\right)^2}{b^2+c^2+c^2+a^2}\le\frac{1}{4}\Sigma\left(\frac{b^2}{b^2+c^2}+\frac{a^2}{c^2+a^2}\right)=\frac{3}{4}\)
\(GT\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Ta có:
\(2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\frac{1}{a^2}+1+\frac{1}{b^2}+1+\frac{1}{c^2}+1\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Cộng vế với vế:
\(3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+3\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=12\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)