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\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
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 \(a=b=c=1\)
ta có: a,b,c>0 mà a+b+c=1 \(\Rightarrow\left(1-a\right)\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a\left(a-b\right)^2\le\left(a-b\right)^2\)
tương tự và cộng theo vế: \(VT\le6\left(ab+bc+ca\right)+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(=2\left(a+b+c\right)^2=2\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
Câu hỏi của nguyen thu phuong - Toán lớp 8 - Học toán với OnlineMath
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
Đặt: \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{xyz}\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta có:
\(S=\frac{\frac{1}{x}}{\sqrt{\frac{1}{y}.\frac{1}{z}\left(1+\frac{1}{x^2}\right)}}+\frac{\frac{1}{y}}{\sqrt{\frac{1}{z}.\frac{1}{x}\left(1+\frac{1}{y^2}\right)}}+\frac{\frac{1}{z}}{\sqrt{\frac{1}{x}.\frac{1}{y}\left(1+\frac{1}{z^2}\right)}}\)
\(=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
\(=\sqrt{\frac{yz}{xy+yz+zx+x^2}}+\sqrt{\frac{zx}{xy+yz+zx+y^2}}+\sqrt{\frac{xy}{xy+yz+zx+z^2}}\)
\(=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\frac{zx}{\left(y+x\right)\left(y+z\right)}}+\sqrt{\frac{xy}{\left(z+x\right)\left(z+y\right)}}\)
\(\le\frac{1}{2}.\left(\frac{y}{x+y}+\frac{z}{x+z}+\frac{z}{y+z}+\frac{x}{x+y}+\frac{x}{z+x}+\frac{y}{z+y}\right)\)
\(=\frac{1}{2}.\left(1+1+1\right)=\frac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=\sqrt{3}\)
\(ab+bc+ca=abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Đặt \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
\(\frac{a}{bc\left(a+1\right)}=\frac{\frac{1}{x}}{\frac{1}{y}\cdot\frac{1}{z}\left(\frac{1}{x}+1\right)}=\frac{xyz}{x\left(x+1\right)}=\frac{yz}{x+1}\)
Tươn tự rồi cộng vế theo vế:
\(A=\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\le\frac{\left(x+y\right)^2}{4\left(z+1\right)}+\frac{\left(y+z\right)^2}{4\left(x+1\right)}+\frac{\left(z+x\right)^2}{4\left(y+1\right)}\)
Đặt \(x+y=p;y+z=q;z+x=r\Rightarrow p+q+r=2\)
\(A\le\Sigma\frac{\left(x+y\right)^2}{4\left(z+1\right)}=\Sigma\frac{\left(x+y\right)^2}{4\left[\left(z+y\right)+\left(z+x\right)\right]}=\frac{p^2}{4\left(q+r\right)}+\frac{r^2}{4\left(p+q\right)}+\frac{q^2}{4\left(p+r\right)}\)
Sau khi đổi biến,cô si thì em ra thế này.Ai đó giúp em với :)
Xét biểu thức \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\)
\(=\frac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(abc+ab+bc+ca\right)+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{4+\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\)(Do \(ab+bc+ca+abc=4\)theo giả thiết)
\(=\frac{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}{\left(ab+bc+ca\right)+4\left(a+b+c\right)+12}=1\)(***)
Với x,y dương ta có 2 bất đẳng thức phụ sau:
\(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)(*)
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)(**)
Áp dụng (*) và (**), ta có:
\(\frac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\frac{1}{a+b+4}=\frac{1}{\left(a+2\right)+\left(b+2\right)}\)
\(\le\frac{1}{4}\left(\frac{1}{a+2}+\frac{1}{b+2}\right)\)(1)
Tương tự ta có: \(\frac{1}{\sqrt{2\left(b^2+c^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{b+2}+\frac{1}{c+2}\right)\)(2)
\(\frac{1}{\sqrt{2\left(c^2+a^2\right)}+4}\le\frac{1}{4}\left(\frac{1}{c+2}+\frac{1}{a+2}\right)\)(3)
Cộng từng vế của các bất đẳng thức (1), (2), (3), ta được:
\(P\le\frac{1}{2}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)=\frac{1}{2}\)(theo (***))
Đẳng thức xảy ra khi \(a=b=c\)
đặt ab=x, bc=y, ac=z
suy ra \(x^3+y^3+z^3=3xyz\)
pt thanh nhân tử \(\left(x+y+z\right)\left(x^2+y^2+z^2-xz-xy-yz\right)=0\)
do x,y,z>0suy ra x+y+z>0
nên suy ra \(x^2+y^2+z^2-xz-yz-xy=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xz-2xy-2yz=0\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
suy ra x=y=z
thế vào pt ta có dpcm
\(ab+bc+ca=abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
\(P=\sum\frac{yz}{x+1}=\sum\frac{yz}{x+x+y+z}=\sum\frac{yz}{x+y+x+z}\le\frac{1}{4}\sum\left(\frac{yz}{x+y}+\frac{yz}{x+z}\right)\)
\(P\le\frac{1}{4}\left(x+y+z\right)=\frac{1}{4}\)
\(P_{max}=\frac{1}{4}\) khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=3\)