Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{a}{bc}+\dfrac{b}{ac}>=2\cdot\sqrt{\dfrac{a}{bc}\cdot\dfrac{b}{ac}}=\dfrac{2}{cc}\)
\(\dfrac{b}{ca}+\dfrac{c}{ab}>=2\cdot\sqrt{\dfrac{bc}{ca\cdot ab}}=\dfrac{2}{a}\)
\(\dfrac{c}{ab}+\dfrac{a}{bc}>=2\cdot\sqrt{\dfrac{a\cdot c}{a\cdot b\cdot c\cdot b}}=\dfrac{2}{b}\)
=>a/bc+b/ac+c/ab>=2(1/a+1/b+1/c)
áp dụng BĐT cô si cho 2 số ta có
\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{bc}{a}.\dfrac{ac}{b}}\)
⇔\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{c^2}=2c\)
TT ta có \(\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\)
\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2b\)
cộng từng vế 3 BĐT trên
\(2\left(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\right)\ge2\left(a+b+c\right)\)
⇔ \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\) (đpcm)
\(a,b,c\ne0\)
\(\dfrac{ac+bc-c^2}{abc}-\dfrac{ab+ac-a^2}{abc}-\dfrac{ab+bc-b^2}{abc}=0\)
\(\Leftrightarrow\dfrac{ac+bc-c^2-ab-ac+a^2-ab-bc+b^2}{abc}=0\)
\(\Leftrightarrow a^2+b^2-c^2-2ab=0\)
\(\Leftrightarrow\left(a-b\right)^2-c^2=0\)
\(\Leftrightarrow\left(a-b-c\right)\left(a-b+c\right)=0\)
\(\Leftrightarrow\left(b+c-a\right)\left(a+c-b\right)=0\) \(\Rightarrow\left[{}\begin{matrix}b+c-a=0\\a+c-b=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{b+c-a}{bc}=0\\\dfrac{a+c-b}{ac}=0\end{matrix}\right.\) (đpcm)
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
Áp dụng bất đẳng thức AM - GM ta ccó :
\(\frac{a}{bc}+\frac{b}{ac}\ge2\sqrt{\frac{a}{bc}.\frac{b}{ac}}=2\sqrt{\frac{1}{c^2}}=\frac{2}{c}\)(1)
\(\frac{b}{ac}+\frac{c}{ab}\ge2\sqrt{\frac{b}{ac}.\frac{c}{ab}}=2\sqrt{\frac{1}{a^2}}=\frac{2}{a}\)(2)
\(\frac{a}{bc}+\frac{c}{ab}\ge2\sqrt{\frac{a}{bc}.\frac{c}{ab}}=2\sqrt{\frac{1}{b^2}}=\frac{2}{b}\)(3)
Cộng vế với vế của (1);(2);(3) lại ta được :
\(\frac{2a}{bc}+\frac{2b}{ac}+\frac{2c}{ab}\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\Leftrightarrow2\left(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)(đpcm)
Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(a^2+b^2+c^2+2ac+2ab+2bc=a^2+b^2+c^2\)
\(ab+bc+ca=0\)
\(ab+bc=-ac\)
\(\left(ab+bc\right)^3=-a^3c^3\)
\(a^3c^3+a^3b^3+b^3c^3+3ab^2c\left(ab+bc\right)=0\)
\(a^3c^3+a^3b^3+b^3c^3=-3ab^2c\left(-ac\right)\)
\(a^3c^3+a^3b^3+b^3c^3=3a^2b^2c^2\)
Ta có:
\(\dfrac{bc}{a^2}+\dfrac{ab}{c^2}+\dfrac{ac}{b^2}=\dfrac{b^3c^3+a^3b^3+a^3c^3}{a^2b^2c^2}=\dfrac{3a^2b^2c^2}{a^2b^2c^2}=3\)
-C/m bằng phép biến đổi tương đương:
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
\(\Leftrightarrow\dfrac{a^2b^2+b^2c^2+a^2c^2}{abc}\ge a+b+c\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge a^2bc+ab^2c+abc^2\)
\(\Leftrightarrow2a^2b^2+2b^2c^2+2c^2a^2-2a^2bc-2ab^2c-2abc^2\ge0\)
\(\Leftrightarrow a^2\left(b^2-2bc+c^2\right)+b^2\left(c^2-2ca+a^2\right)+c^2\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow a^2\left(b-c\right)^2+b^2\left(c-a\right)^2+c^2\left(a-b\right)^2\ge0\) (luôn đúng)
-Dấu "=" xảy ra khi \(a=b=c\)