a) Áp dụng BĐT Cô si cho 2 số dương ta có :

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}\ge2b\)

b) \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)

CMTT như câu a ta đc :

\(\frac{ab}{c}+\frac{bc}{a}\ge2b;\frac{ab}{c}+\frac{ca}{b}\ge2a;\frac{bc}{a}+\frac{ac}{b}\ge2c\)

Do đó : \(\frac{ab}{c}+\frac{bc}{a}+\frac{ab}{c}+\frac{ca}{b}+\frac{bc}{a}+\frac{ca}{b}\ge2a+2b+2c\)

\(\Rightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\left(đpcm\right)\)

a. Áp dung BĐT AM-GM:

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}=2\sqrt{b^2}=2b\)

b. Áp dung BĐT AM-GM:

\(\frac{ab}{c}+\frac{bc}{a}\ge2b\)

\(\frac{bc}{a}+\frac{ca}{b}\ge2c\)

\(\frac{ca}{b}+\frac{ab}{c}\ge2a\)

\(\Rightarrow2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)

Xảy ra đẳng thức khi \(a=b=c>0\)

a/ \(\frac{ab}{c}+\frac{bc}{a}\ge2b\)

\(\Leftrightarrow a^2b+bc^2\ge2abc\)

\(\Leftrightarrow a^2b+bc^2-2abc\ge0\)

\(\Leftrightarrow\left(a\sqrt{b}-c\sqrt{b}\right)^2\ge0\)(đúng)

\(\RightarrowĐPCM\)

b/ Áp dụng câu a ta có

\(\frac{ab}{c}+\frac{bc}{a}\ge2b\)

\(\frac{ab}{c}+\frac{ca}{b}\ge2a\)

\(\frac{bc}{a}+\frac{ac}{b}\ge2c\)

Cộng 3 cái đó vế theo vế được

\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)

a) Áp dụng bất đẳng thức AM-GM ta có ngay :

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2\sqrt{\frac{ab^2c}{ac}}=2\sqrt{b^2}=2\left|b\right|=2b\)( do b > 0 )

=> đpcm

Đẳng thức xảy ra <=> a = b = c

b) Áp dụng bất đẳng thức AM-GM ta có :

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2b\)(1) ( như a) đấy :)) )

tương tự : \(\frac{bc}{a}+\frac{ca}{b}\ge2c\)(2) ; \(\frac{ab}{c}+\frac{ca}{b}\ge2a\)(3)

Cộng (1), (2), (3) theo vế ta có đpcm

Đẳng thức xảy ra <=> a = b = c

c) \(\frac{a^3+b^3}{2ab}+\frac{b^3+c^3}{2bc}+\frac{c^3+a^3}{2ca}\)

\(=\frac{a^3}{2ab}+\frac{b^3}{2ab}+\frac{b^3}{2bc}+\frac{c^3}{2bc}+\frac{c^3}{2ca}+\frac{a^3}{2ca}\)

\(=\frac{a^2}{2b}+\frac{b^2}{2a}+\frac{b^2}{2c}+\frac{c^2}{2b}+\frac{c^2}{2a}+\frac{a^2}{2c}\)(I)

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

\(\left(I\right)\ge\frac{\left(a+b+b+c+c+a\right)^2}{2b+2a+2c+2b+2a+2c}=\frac{\left[2\left(a+b+c\right)\right]^2}{4\left(a+b+c\right)}=\frac{4\left(a+b+c\right)^2}{4\left(a+b+c\right)}=a+b+c\)

hay \(\frac{a^3+b^3}{2ab}+\frac{b^3+c^3}{2bc}+\frac{c^3+a^3}{2ca}\ge a+b+c\)(đpcm)

Đẳng thức xảy ra <=> a = b = c

Câu b nhá mn

quá dễ BĐT_AM-GM sẽ cân tất cả

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{abbc}{ac}}=2\sqrt{b^2}=2b;tươngtự:\left\{{}\begin{matrix}\frac{bc}{a}+\frac{ac}{b}\ge2\sqrt{\frac{abc^2}{ab}}=2c\\\frac{ac}{b}+\frac{ab}{c}\ge2\sqrt{\frac{a^2bc}{bc}}=2a.Cộngvếtheovếtađược:2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\end{matrix}\right.\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\left(\text{đpcm}\right)\)

Ta có :

\(\dfrac{a}{b} = \dfrac{b}{c} => \dfrac{a^2}{b^2} =\dfrac{b^2}{c^2} \) \(= \dfrac{a.b}{b.c} = \dfrac{a}{c} (1)\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\dfrac{a^2}{b^2} =\dfrac{b^2}{c^2} =\dfrac{a^2+b^2}{b^2+c^2} (2)\)

Từ (1) và (2) => \(\dfrac{a^2+b^2}{b^2+c^2} = \dfrac{a}{c}\) (đpcm)

ab +bc+ca =0 nên 1/a+ 1/b+ 1/c = 0 (chia cả 2 vế cho abc)

Ta có: S = b+c / a +c+a /b + a+b /c

                = a+b+c /a +a+b+c /b +a+b+c /c -3

                = (a+b+c)(1/a +1/b+1/c)-3

                = 0-3 (vì 1/a+ 1/b+ 1/c = 0)

                = -3

Vậy S = -3

Chúc bạn học tốt.

nhanh lắm Phạm Trọng An Nam

trả lời

dùng bất đẳng thức cosi cho 2 số ko âm

sử dụng cộng mỗi cặp trên

đc 3 cặp

cộng lại là ra

ta có : \(\frac{ab}{c}+\frac{ca}{b}\ge2a;\frac{bc}{a}+\frac{ca}{b}\ge2c\)

Do đó : \(\frac{ab}{c}+\frac{bc}{a}+\frac{ab}{c}+\frac{ca}{b}+\frac{bc}{a}+\frac{ca}{b}\ge2b+2a+2c\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)

Ta có : \(\frac{a^2-bc}{a}+\frac{b^2-ac}{b}+\frac{c^2-ab}{c}=0\)

=> \(a-\frac{bc}{a}+b-\frac{ac}{b}+c-\frac{ab}{c}=0\)

=> \(a+b+c=\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\)

=> \(a+b+c=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

=> \(\frac{a+b+c}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

=> \(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

=> \(\frac{2}{bc}+\frac{2}{ac}+\frac{2}{ab}=\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\)

=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{bc}-\frac{2}{ac}-\frac{2}{ac}=0\)

=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{a^2}-\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{b^2}-\frac{1}{bc}+\frac{1}{c^2}\right)=0\)

=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{a}-\frac{1}{c}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2=0\)

=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{a}-\frac{1}{c}=0\\\frac{1}{b}-\frac{1}{c}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}\\\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{c}\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\left(\text{đpcm}\right)\)