Ta có:

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

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow a=-b;b=-c;c=-a\)

Với \(a=-b\)ta có

\(a^3+b^3+c^3=1\)

\(\Leftrightarrow c^3=1\)

\(\Leftrightarrow c=1\)

Thì ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a}-\frac{1}{a}+\frac{1}{c}=1\)

Tương tự cho 2 trường hợp còn lại được ĐPCM

THƯA CHỊ BÀI NÀY LÀ SAO AK, E HỌC LỚP 5 ** BIK BÀI NÀY NHÉ ~_~ !!!!!!!!!!!

vậy em giải giùm chị nhé

đặt \(\sqrt{\frac{ab}{c}}=x;\sqrt{\frac{bc}{a}}=y;\sqrt{\frac{ca}{b}}=z\Rightarrow xy+yz+zx=1\)

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

\(=\frac{\frac{ab}{c}}{\frac{ab}{c}+1}+\frac{\frac{bc}{a}}{\frac{bc}{a}+1}+\frac{\frac{ca}{b}}{\frac{ca}{b}+1}=\frac{x^2}{x^2+1}+\frac{y^2}{y^2+1}+\frac{z^2}{z^2+1}\)

\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\frac{\left(x+y+z\right)^2}{3}}=\frac{3}{4}\left(Q.E.D\right)\)

Ta có :(a+b-c)² \(\ge\) 0

<=>a²+b²+c² \(\ge\) 2(bc-ab+ac)

<=>\(\frac{5}{3}\ge\) 2(bc-ab+ac)

<=>bc+ac-ab \(\le\frac{5}{6}< 1\)

<=>\(\frac{bc+ac-ab}{abc}< \frac{1}{abc}\) (vì a,b,c>0 nên chia cả 2 vế cho abc)

<=>\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< 1\) (đpcm)

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

\(< =\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{1}{9}\left(\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac{1}{9}\left(\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)\)(bđt svacxo)

\(=\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)=\frac{1}{9}\cdot3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

\(=\frac{1}{9}\cdot3\cdot\frac{1}{3}=\frac{1}{9}\cdot1=\frac{1}{9}\)

\(\Rightarrow\frac{1}{2a^2+b^2}+\frac{1}{2b^2+c^2}+\frac{1}{2c^2+a^2}< =\frac{1}{9}\)(đpcm)

dấu = xảy ra khi \(\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}=\frac{1}{9}\Rightarrow a=b=c=3\)

ta có:\(\left(a+2b\right)^2=\left(1.a+\sqrt{2}.\sqrt{2}b\right)^2\le\left(1+2\right)\left(a^2+2b^2\right)\)( bđt bunhiacopxki)

\(\left(a+2b\right)^2\le3.3c^2=9c^2\)→\(a+2b\le3c\)

lại có:\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\)

dấu = xảyra khi.... a+2b²=3c²(:v)

ra nguyên những bài khó hiểu hết là chết liền

hoc lop may vay bai kho qua ai ma giai duoc