\(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\)

\(S=\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{b}{a}+\frac{a}{b}\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\hept{\begin{cases}\frac{a}{c}+\frac{c}{a}\ge2\sqrt{\frac{ac}{ca}}=2\\\frac{b}{c}+\frac{c}{b}\ge2\sqrt{\frac{bc}{cb}}=2\\\frac{b}{a}+\frac{a}{b}\ge2\sqrt{\frac{ab}{ba}}=2\end{cases}}\)

\(\Rightarrow\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{b}{a}+\frac{a}{b}\right)\ge2+2+2=6\)

\(\Leftrightarrow\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge6\)

\(\Leftrightarrow S\ge6\left(đpcm\right)\)

\(\Rightarrow S_{min}=6\)

Dấu " = " xảy ra khi \(a=b=c\)

Chúc bạn học tốt !!!

\(\left\{{}\begin{matrix}s_1=\dfrac{b}{a}x+\dfrac{c}{a}z\\s_2=\dfrac{a}{b}x+\dfrac{c}{b}y\\s_3=\dfrac{a}{c}z+\dfrac{b}{c}y\\x+y+z=5\end{matrix}\right.\) \(\left\{{}\begin{matrix}s_1+s_2+s_3=\left(\dfrac{b}{a}+\dfrac{a}{b}\right)x+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)y+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)z\\a,b,c\in N\left(sao\right)\\\dfrac{b}{a}+\dfrac{a}{b}\ge2;\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\ge2;\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge2\\x+y+z=5\end{matrix}\right.\)

\(s_1+s_2+s_3\ge2x+2y+2z\ge2\left(x+y+z\right)=2.5=10\)

1.VT= \(\dfrac{x}{z}+\dfrac{y}{z}+\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{x}{y}=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)\)

Áp dụng BĐT Cô-si cho 2 số dương, ta có:

\(\dfrac{x}{y}+\dfrac{y}{x}\)≥ 2\(\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}\)=2; tương tự \(\dfrac{x}{z}+\dfrac{z}{x}\)≥2; \(\dfrac{y}{z}+\dfrac{z}{y}\)≥2.

Cộng 3 BĐT trên, ta được đpcm.

2.Đặt b+c-a= x, a+c-b= y, a+b-c= z. Khi đó x,y,z>0.

2a= y+z; 2b= x+z; 2c= x+y. Khi đó bđt cần chứng minh trở thành:

\(\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{z+x}{y}\)≥6.

Theo bài 1 bđt luôn đúng

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{a}+\frac{1}{b}\right)(a+b)\ge (1+1)^2\)

\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}\)

\(\Rightarrow \frac{c}{a}+\frac{c}{b}\geq \frac{4c}{a+b}\)

Hoàn toàn tương tự: \(\frac{a}{b}+\frac{a}{c}\geq \frac{4a}{b+c}; \frac{b}{a}+\frac{b}{c}\geq \frac{4b}{a+c}\)

Cộng theo vế các BĐT thu được:

\(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq 4\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\) (đpcm)

Dấu bằng xảy ra khi $a=b=c$

Ta có: \(\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{2}{a}\)

\(\Leftrightarrow\dfrac{b+c}{bc}=\dfrac{2}{a}\Leftrightarrow ab+ac=2bc\)

\(\dfrac{a+b}{a-b}+\dfrac{a+c}{a-c}=\dfrac{a^2-ac+ab-bc+a^2+ac-ab-bc}{a^2-ac-ab+bc}\)

\(=\dfrac{2a^2-2bc}{a^2-2bc+bc}=\dfrac{2a^2-2bc}{a^2-bc}=2\)

\(\Rightarrowđpcm\)

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

BPT?

a/d vào công thức a^3+b^3+b^3=3abc( khi a+b+c=0)

ta đc 1/a+1/b+1/c=0

=> (1/a)^3+(1/b)^3+(1/c)^3=3. (1/abc)

lại có S=\(\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)

=abc (\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\))

=3.\(\dfrac{abc}{abc}\)=1

chúc bạn học tốt ^ ^

Dễ CM : nếu x+y+z=0 thì x^3+y^3+z^3=3xyz

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)

\(S=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\\ =abc.\dfrac{1}{abc}=1\)