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Với a,b,c > 0
Áp dụng bđt cosi cho 2 số dương \(\frac{a^2}{b^2}\)và \(\frac{b^2}{c^2}\), ta có:
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{b^2}\cdot\frac{b^2}{c^2}}=2\frac{a}{c}\) (1)
CMTT: \(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge2\frac{c}{b}\)(2)
\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\frac{b}{a}\)(3)
Từ (1), (2) và (3) cộng vế theo vế:
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{a^2}{b^2}+\frac{c^2}{a^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\frac{a}{c}+2\frac{c}{b}+2\frac{b}{a}\)
<=> \(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\)
<=> \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\)
Ta có : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\) (1)
Ta có : a+b+c khác 0
do nếu a+b+c=0=>\(\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}=1\)=>-3=1(Vô lí)
do a+b+c khác 0 nên ta nhân (a+b+c) vào (1)
=>\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
=>\(\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(c+a\right)}{c+a}+\frac{c\left(a+b\right)+c^2}{a+b}=a+b+c\)
=>\(\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)(ĐPCM)
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
\(\Leftrightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\frac{ab}{c+a}+\frac{ac}{a+b}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{ac}{b+c}+\frac{bc}{c+a}=a+b+c\)
\(\Leftrightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\frac{ab+bc}{c+a}+\frac{ac+bc}{a+b}+\frac{ab+ac}{b+c}=a+b+c\)
\(\Leftrightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+a+b+c=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
Bạn ơi , bao giờ giáo viên của bạn chữa cho bạn bài này thì cho mình xin lời giải nhé , mình cám ơn ạ !
\(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}\)
\(\frac{b^2}{a^2+c^2}-\frac{b}{a+c}=\frac{ab\left(b-a\right)+bc\left(b-c\right)}{\left(a^2+c^2\right)\left(a+c\right)}\)
\(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{ac\left(c-a\right)+bc\left(c-b\right)}{\left(a^2+b^2\right)\left(a+b\right)}\)
Cộng các vế ta có:
\(\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(=ab\left(a-b\right)\left[\frac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\frac{1}{\left(a^2+c^2\right)\left(a+c\right)}\right]\)\(+ac\left(a-c\right)\left[\frac{1}{\left(b^2+c^2\right)\left(b+c\right)}-\frac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\)
\(+bc\left(b-c\right)\left[\frac{1}{\left(a^2+c^2\right)\left(a+c\right)+}-\frac{1}{\left(a^2+b^2\right)\left(a+b\right)}\right]\)
Giả sử \(a\ge b\ge c>0\)thì
\(\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)>0\)
=> \(\frac{a^2}{b^2+c^2}+\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+b^2}\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Dấu " = " xảy ra <=> a=b=c
em học lớp 5 nên k hiểu được bài lớp 8 nhưng cứ comments,hi
a)có \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)=\left(a+b\right)\left(\frac{a+b}{ab}\right)\)\(=\frac{\left(a+b\right)^2}{ab}\Rightarrow\frac{\left(a+b\right)^2}{ab}-4=\frac{\left(a+b\right)^2-4ab}{ab}=\frac{\left(a-b\right)^2}{ab}\)\(\ge0\forall a;b>0\)
\(\Rightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)dấu''=''xảy ra khi a=b
b)B=\(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}\)
=\(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\)mà ta có \(\frac{x}{y}+\frac{y}{x}-2=\frac{\left(x-y\right)^2}{xy}\ge0\forall x;y>0\)
\(\Rightarrow\frac{x}{y}+\frac{y}{x}\ge2\forall x;y>0\)áp dụng bđt trên ta có B\(\ge\)2+2+2=6
dấu ''=''xảy ra khi x=y=z
1)Áp dụng Bđt Am-Gm \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\)
2)Áp dụng Am-Gm \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;b^2+c^2\ge2bc;a^2+c^2\ge2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
=>ĐPcm
3)(a+b+c)2\(\ge\)3(ab+bc+ca)
=>a2+b2+c2+2ab+2bc+2ca\(\ge\)3ab+3bc+3ca
=>a2+b2+c2-ab-bc-ca\(\ge\)0
=>2a2+2b2+2c2-2ab-2bc-2ca\(\ge\)0
=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ac+a2)\(\ge\)0
=>(a-b)2+(b-c)2+(c-a)2\(\ge\)0
4)đề đúng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)