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Ồ sorry bạn nhiều, chỗ đấy bị lỗi kĩ thuật rồi, mình sửa lại nhé :
\(M\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
Lại có : \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt{a^3b^3c^3}}{2}=\frac{3}{2}\)
Do đó : \(M\ge\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Ta có : \(\frac{1}{a^3\left(b+c\right)}=\frac{\frac{1}{a^2}}{a\left(b+c\right)}=\frac{\left(\frac{1}{a}\right)^2}{a\left(b+c\right)}\)
Tương tự : \(\frac{1}{b^3\left(a+c\right)}=\frac{\left(\frac{1}{b}\right)^2}{b\left(a+c\right)}\) , \(\frac{1}{c^3\left(a+b\right)}=\frac{\left(\frac{1}{c}\right)^2}{c\left(a+b\right)}\)
Ta thấy : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Áp dụng BĐT Svacxo ta có :
\(M=\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^2\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\) \(\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Vâỵ \(M_{min}=\frac{3}{2}\) tại \(a=b=c=1\)
Áp dụng BĐT Svac - xơ:
\(T=\frac{a}{a^2+8bc}+\frac{b}{b^2+8ca}+\frac{c}{c^2+8ab}\)
\(=\frac{a^2}{a^3+8abc}+\frac{b^2}{b^3+8abc}+\frac{c^2}{c^3+8abc}\)\(\ge\frac{\left(a+b+c\right)^2}{a^3+b^3+c^3+24abc}\)
Ta lại có: \(\left(a+b+c\right)^3=a^3+b^3+c^3+\)\(3\left(a+b+c\right)\left(ab+bc+ca\right)-3abc\)
\(\ge a^3+b^3+c^3+27\sqrt[3]{abc}.\sqrt[3]{\left(abc\right)^2}-3abc=\)\(a^3+b^3+c^3+24abc\)
Lúc đó: \(T\ge\frac{1}{a+b+c}=1\)
(Dấu "="\(\Leftrightarrow a=b=c=\frac{1}{3}\))
Áp dụng BĐT AM-GM (Cô si): \(A\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
\(=3\sqrt[3]{\frac{1}{a\left(b+c\right).b\left(c+a\right).c\left(a+b\right)}}=\frac{3}{\sqrt[3]{\left(ab+ca\right)\left(bc+ab\right)\left(ca+bc\right)}}\)
\(\ge\frac{9}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
P/s: Check giúp em xem có ngược dấu không:v
Cach khac
Dat \(\left(ab;bc;ca\right)\rightarrow\left(x;y;z\right)\)
\(\Rightarrow\hept{\begin{cases}x+y+z=3\\x^2+y^2+z^2\ge3\\xyz\le1\end{cases}}\)
Ta co:
\(A=\frac{1}{ab+b^2}+\frac{1}{bc+c^2}+\frac{1}{ca+a^2}\)
\(=\frac{1}{x+\frac{xy}{z}}+\frac{1}{y+\frac{yz}{x}}+\frac{1}{z+\frac{zx}{y}}\ge\frac{9}{3+xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)}\ge\frac{9}{3+3}=\frac{3}{2}\)
Dau '=' xay ra khi \(a=b=c=1\)
Vay \(A_{min}=\frac{3}{2}\)khi \(a=b=c=1\)
bạn vào đây tham khảo nè
Câu hỏi của Tuấn Anh - Toán lớp 9 | Học trực tuyến
hơi lằng nhằng 1 chút
\(P=\frac{a}{\sqrt{a+2c}+1}+\frac{b}{\sqrt{b+2a}+1}+\frac{c}{\sqrt{c+2b}+1}\)
áp dụng cô si ta có:
\(\left(\sqrt{a+2c}+1\right)^2\le2\left(a+2c+1\right)=2\left(2a+b+3c\right)\)
tương tự \(\Rightarrow P\ge\frac{a}{\sqrt{2\left(2a+b+3c\right)}}+\frac{b}{\sqrt{2\left(2b+c+3a\right)}}+\frac{c}{\sqrt{2\left(2c+a+3b\right)}}\)
mà \(\sqrt{2\left(2a+b+3c\right)}\le\frac{2a+b+3c+2}{2}=\frac{4a+3b+5c}{2}\)
\(\Rightarrow P\ge\frac{2a}{4a+3b+5c}+\frac{2b}{4b+3c+5a}+\frac{2c}{4c+3a+5b}\)
\(=\frac{2a^2}{4a^2+3ab+5ac}+\frac{2b^2}{4b^2+3bc+5ab}+\frac{2c^2}{4c^2+3ac+5bc}\ge\frac{2\left(a+b+c\right)^2}{4\left(a+b+c\right)^2}=\frac{1}{2}\)
a + b + c= 1 \(\Rightarrow\)1 - a = b + c > 0
Tương tự : 1 - b > 0 ; 1 - c > 0
Mà 1 + a = 1 + ( 1 - b - c ) = ( 1- b ) + ( 1 - c ) \(\ge\)\(2\sqrt{\left(1-b\right)\left(1-c\right)}\)
Tương tự : \(1+b\ge2\sqrt{\left(1-a\right)\left(1-c\right)}\); \(1+c\ge2\sqrt{\left(1-a\right)\left(1-b\right)}\)
\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge8\sqrt{\left(1-a\right)^2\left(1-b\right)^2\left(1-c\right)^2}=8\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
\(\Rightarrow A=\frac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\ge8\)
Dấu " = : xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Vậy GTNN của A là 8 \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Cách khác:
\(A=\frac{\left[\left(a+b\right)+\left(a+c\right)\right]\left[\left(b+c\right)+\left(b+a\right)\right]\left[\left(c+a\right)+\left(c+b\right)\right]}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Áp dụng BĐT Cô si cho 2 số ta được:
\(A\ge\frac{8\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=8\)
"=" <=> a = b = c = 1/3
Kết luận..
\(A=\dfrac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}=\dfrac{\left(a+b+c+a\right)\left(b+a+b+c\right)\left(c+a+b+c\right)}{\left(b+c\right)\left(a+c\right)\left(a+b\right)}\)
\(A\ge\dfrac{2\sqrt{\left(a+b\right)\left(c+a\right)}.2\sqrt{\left(a+b\right)\left(b+c\right)}.2\sqrt{\left(a+c\right)\left(b+c\right)}}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=8\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)