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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)
Áp dụng BĐT Cauchy-Schwarz :
\(\dfrac{ac}{b}+\dfrac{ab}{c}\ge2.\sqrt{\dfrac{ac}{b}.\dfrac{ab}{c}}=2.\sqrt{a^2}=2a\\ \dfrac{ab}{c}+\dfrac{bc}{a}\ge2\sqrt{\dfrac{ab}{c}.\dfrac{bc}{a}}=2.\sqrt{b^2}=2b\\ \dfrac{ac}{b}+\dfrac{bc}{a}\ge2.\sqrt{\dfrac{ac}{b}.\dfrac{bc}{a}}=2.\sqrt{c^2}=2c\\ \Rightarrow2\left(\dfrac{ac}{b}+\dfrac{ab}{c}+\dfrac{bc}{a}\right)\ge2\left(a+b+c\right)\\ \Rightarrow\dfrac{ac}{b}+\dfrac{ab}{c}+\dfrac{bc}{a}\ge a+b+c\)
AM-GM ngược dấu như sau:
\(\dfrac{a^3}{a^2+ab+b^2}=a-\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\dfrac{ab\left(a+b\right)}{3ab}=\dfrac{2a-b}{3}\)
Tương tự ta cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{b^2+bc+c^2}\ge\dfrac{2b-c}{3};\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{2c-a}{3}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a-b}{3}+\dfrac{2b-c}{3}+\dfrac{2c-a}{3}=\dfrac{a+b+c}{3}=VP\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+a^2b+ab^2+b^2c+bc^2+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^3+b^3+c^3+ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)}\)
Dễ thấy :
\(a^{3}+b^{3}+c^{3}+ab(b+c)+bc(b+c)+ca(c+a)=(a^{2}+ b^{2}+c^{2})(a+b+c)\)
\(\Rightarrow VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
Vậy cần chứng minh
\(\dfrac{a^2+b^2+c^2}{a+b+c}\ge\dfrac{a+b+c}{3}\Leftrightarrow\left(a+b+c\right)^2\ge3\left(a^2+b^2+c^2\right)\) (luôn đúng)
-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 \(a=b=c\)
e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM
Áp dụng BĐT Cauchy ta có
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge a\)
\(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\)
\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c
Làm tắt vài chỗ thông cảm
Câu b,
Ta có BĐT Cauchy \(a^2+b^2\ge2ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)
\(\Rightarrow\dfrac{ab}{a+b}\le\dfrac{\left(a+b\right)^2}{4\left(a+b\right)}=\dfrac{a+b}{4}\)
Tương tự \(\dfrac{bc}{b+c}\le\dfrac{b+c}{4}\)
\(\dfrac{ac}{a+c}\le\dfrac{a+c}{4}\)
Cộng theo vế ta đc \(VT\le\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c
\(\dfrac{a}{bc}+\dfrac{b}{ac}>=2\cdot\sqrt{\dfrac{a}{bc}\cdot\dfrac{b}{ac}}=\dfrac{2}{cc}\)
\(\dfrac{b}{ca}+\dfrac{c}{ab}>=2\cdot\sqrt{\dfrac{bc}{ca\cdot ab}}=\dfrac{2}{a}\)
\(\dfrac{c}{ab}+\dfrac{a}{bc}>=2\cdot\sqrt{\dfrac{a\cdot c}{a\cdot b\cdot c\cdot b}}=\dfrac{2}{b}\)
=>a/bc+b/ac+c/ab>=2(1/a+1/b+1/c)
Bài 1:
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\) với a,b,c > 0
Áp dụng BĐT Chauchy cho 2 số không âm, ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge c\sqrt{\dfrac{b}{a}.\dfrac{a}{b}}=2c\)
\(\dfrac{ac}{b}+\dfrac{ab}{c}=a\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\ge a\sqrt{\dfrac{c}{b}.\dfrac{b}{c}}=2a\)
\(\dfrac{ab}{c}+\dfrac{bc}{a}=b\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge b\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2b\)
Cộng vế theo vế ta được:
\(2\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
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)