* Áp dụng BĐT \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với $x,y>0$ vào bài toán có :

\(\dfrac{1}{4}\cdot\left(\dfrac{4}{a+b}\right)\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(\dfrac{1}{4}\left(\dfrac{4}{b+c}\right)\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\dfrac{1}{4}\left(\dfrac{4}{c+a}\right)\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)

Cộng vế với vế các BĐT có :

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

\(\dfrac{1}{a^2+a+1}\ge\dfrac{1}{a^2+\dfrac{a^2+1}{2}+1}=\dfrac{2}{3}.\dfrac{1}{a^2+1}=\dfrac{2}{3}\left(1-\dfrac{a^2}{a^2+1}\right)\ge\dfrac{2}{3}\left(1-\dfrac{a}{2}\right)\)

Tương tự và cộng lại: \(VT\ge\dfrac{2}{3}\left(3-\dfrac{a+b+c}{2}\right)=\dfrac{2}{3}.\dfrac{3}{2}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Lời giải:

Ta có

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

\(=a+b+c-(a+b+c)=0\)

Ta có đpcm

Câu 1

\(a+b\ge2\sqrt{ab}\Leftrightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\\ \Leftrightarrow N=ab+\dfrac{1}{16ab}+\dfrac{15}{16ab}\ge2\sqrt{\dfrac{1}{16}}+\dfrac{15}{4\left(a+b\right)^2}\ge\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\)

Dấu \("="\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 2:

\(P=a+\dfrac{1}{a}+2b+\dfrac{8}{b}+3c+\dfrac{27}{c}+4\left(a+b+c\right)\\ P\ge2\sqrt{1}+2\sqrt{16}+2\sqrt{81}+4\cdot6=2+8+18+4=32\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=3\end{matrix}\right.\)

Câu 3: Cho a,b,c là các số thuộc đoạn [ -1;2 ] thõa mãn \(a^2+b^2+c^2=6.\) CMR : \(a+b+c>0\) - Hoc24

Ùi mình làm theo kiểu khác thử :V, nhưng có hơi hướng giống và bổ sung :D

Câu 2 : a,b,c > 0. CM : \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

Giải :

C1 : Áp dụng bất đẳng thức Cauchy - Schwarz dạng Engel ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\left(ĐPCM\right)\)

Đẳng thức xảy ra khi \(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\).

C2 : Đầy đủ hơn với cách giải đúng của bạn Hoàng Thiên Di :

Áp dụng BĐT AM-GM cho 3 số dương (sgk là cosi :v)

\(a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+1+1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

\(\ge3+2+2+2=9\left(ĐPCM\right)\)

Câu 3 : a,b,c > 0. CM : \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)

Giải :

\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)

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

\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge6\)

Theo bất đẳng thức Cosi : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{xy}{yx}}=2\)

Thay vào các vế được : \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\sqrt{1}=2\)

\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\sqrt{1}=2\)

\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\sqrt{1}=2\)

\(\Leftrightarrow2+2+2\ge6\) (đúng)

BĐT được c/m.

xem lại đề

a=b=c=1 =>3<=2

Bài 3 :

Đặt A= \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}=\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}\)+\(\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\)

=\(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\) (*)

Xét BĐT : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\) , với x,y >0

Áp dụng AM-GM => \(\dfrac{x}{y}+\dfrac{y}{x}\)\(\ge2\sqrt{\dfrac{xy}{xy}}=2\)

Thay vào (*) => A \(\ge2+2+2=6\)

Hay A\(\ge6\left(đpcm\right)\)

b) \(\dfrac{1}{3a+2b+c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{36}\left(\dfrac{3}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)

Tương tự cho 2 cái kia rồi cộng lại

\(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}.16=\dfrac{8}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow\) ... \(\Leftrightarrow a=b=c=\dfrac{3}{16}\)

Ta co BDT :\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\forall a,b\in R^+\)

Tuong tu cho 2 BDT con lai ta cung co:

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c};\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+c}\)

Cong theo ve 3 BDT tren ta co

\(VT=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}=VP\)

Dau "=" xay ra khi \(a=b=c\)