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Từ giả thiết \(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Khi đó \(\frac{x}{1+x^2}=\frac{\frac{1}{x}}{\frac{1}{x^2}+1}=\frac{\frac{1}{x}}{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)
Tương tự cho 2 cái còn lại ta có: \(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)
\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)
Suy ra \(VT=\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Đpcm
Trả lời
Từ giả thiết x+y+z=xyz <=> 1/xy + 1/yz + 1/zx = 1
Khi đó: x/1+x2 = \(\frac{1}{\frac{x}{\left(\frac{1}{z}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}}\)\(=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)
Tương tự cho 2 cái còn lại ta có:\(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)
\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)
Suy ra VT=\(\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)\(=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
ĐPCM
Ta có:\(\frac{x}{1+x^2}=\frac{xyz}{yz+x^2yz}=\frac{xyz}{yz+x\left(xyz\right)}=\frac{xyz}{yz+x\left(x+y+z\right)}=\frac{xyz}{yz+x^2+xy+xz}=\frac{xyz}{y\left(x+z\right)+x\left(x+z\right)}\)
\(=\frac{xyz}{\left(x+z\right)\left(y+x\right)}\)
Chứng minh tương tự : \(\frac{2y}{1+y^2}=\frac{2xyz}{\left(y+z\right)\left(y+x\right)}\)
\(\frac{3z}{1+z^2}=\frac{3xyz}{\left(x+z\right)\left(x+y\right)}\)
Khi đó VT \(=\frac{xyz}{\left(x+z\right)\left(y+x\right)}+\frac{2xyz}{\left(y+z\right)\left(y+x\right)}+\frac{3xyz}{\left(x+z\right)\left(z+y\right)}\)
\(=\frac{xyz\left[y+z+2\left(z+x\right)+3\left(x+y\right)\right]}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(đpcm\right)\)
( mình đang vội nên làm hơi tắt mong bạn thông cảm )
Vì xyz = 1 nên ta có thể đặt \(x=\frac{a^2}{bc};y=\frac{b^2}{ac};z=\frac{c^2}{ab}\left(a,b,c>0,a^2\ne bc,b^2\ne ac,c^2\ne ab\right)\)
Khi đó bất đẳng thức tương đương với
\(\frac{a^4}{\left(a^2-bc\right)^2}+\frac{b^4}{\left(b^2-ac\right)^2}+\frac{c^4}{\left(c^2-ab\right)^2}\ge1\)
Mà ta có
\(\frac{a^4}{\left(a^2-bc\right)^2}+\frac{b^4}{\left(b^2-ac\right)^2}+\frac{c^4}{\left(c^2-ab\right)^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2-bc\right)^2+\left(b^2-ab\right)^2+\left(c^2-ab\right)^2}\)
Ta cần chứng minh
\(\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2-bc\right)^2+\left(b^2-ab\right)^2+\left(c^2-ab\right)^2}\ge1\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\left(a^2-bc\right)^2+\left(b^2-ab\right)^2+\left(c^2-ab\right)^2\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge0\left(đúng\right)\)
Vậy ta có điều phải chứng minh
Do x, y, z khác 1 và thỏa mãn xyz = 1 nên ta có thế đặt: \(x=\frac{a^2}{bc};y=\frac{b^2}{ca};z=\frac{c^2}{ab}\)
với \(\left(a^2-bc\right)\left(b^2-ca\right)\left(c^2-ab\right)\ne0\)
Khi đó BĐT cần chứng minh được viết lại như sau:
\(\frac{a^4}{\left(a^2-bc\right)^2}+\frac{b^4}{\left(b^2-ca\right)^2}+\frac{c^4}{\left(c^2-ab\right)^2}\ge1\)
Áp dụng BĐT Bunhiacopxki ta có: \(\left[\text{∑}_{cyc}\left(a^2-bc\right)^2\right]\left[\text{∑}_{cyc}\frac{a^4}{\left(a^2-bc\right)^2}\right]\ge\left(a^2+b^2+c^2\right)^2\)
\(\Rightarrow\text{∑}_{cyc}\frac{a^4}{\left(a^2-bc\right)^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2-bc\right)^2+\left(b^2-ca\right)^2+\left(c^2-ab\right)^2}\)
Đến đây, ta cần chứng minh: \(\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2-bc\right)^2+\left(b^2-ca\right)^2+\left(c^2-ab\right)^2}\ge1\left(^∗\right)\)
Thật vậy. \(\left(^∗\right)\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\left(a^2-bc\right)^2+\left(b^2-ca\right)^2+\left(c^2-ab\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\ge a^4+b^4+c^4\)\(+\left(a^2b^2+b^2c^2+c^2a^2\right)-2\left(a^2bc+ab^2c+abc^2\right)\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2\left(a^2bc+2ab^2c+2abc^2\right)\ge0\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge0\)*đúng*
Vậy bất đẳng thức được chứng minh.
Vì xyz=1 nên x,y,z \(\ne\)0. Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\) thì ta có: \(abc=1\) và \(a,b,c\ne0,1\)
Khi đó BĐT cần chứng minh trở thành
\(\frac{1}{\left(1-a\right)^2}+\frac{1}{\left(1-b\right)^2}+\frac{1}{\left(1-c\right)^2}\ge1\Leftrightarrow\left(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\right)^2\)
\(-2\left[\frac{1}{\left(1-a\right)\left(1-b\right)}+\frac{1}{\left(1-b\right)\left(1-c\right)}+\frac{1}{\left(1-c\right)\left(1-a\right)}\right]\ge1\)
\(\Leftrightarrow\left[\frac{32\left(a+b+c\right)+ab+bc+ca}{ab+bc+ca-\left(a+b+c\right)}\right]^2-2\left[\frac{3-\left(a+b+c\right)}{ab+bc+ca+ca-\left(a+b+c\right)}\right]\ge1\)
\(\Leftrightarrow\left[1+\frac{3-\left(a+b+c\right)}{ab+bc+ca-\left(a+b+c\right)}\right]^2-2\left[\frac{3-\left(a+b+c\right)}{ab+bc+ca-\left(a+b+c\right)}\right]\ge1\)
\(\Leftrightarrow1+\left[\frac{3-\left(a+b+c\right)}{ab+bc+ca-\left(a+b+c\right)}\right]\ge1\)
We have:
\(A=\Sigma_{cyc}\frac{1}{3xy+3zx+x+y+z}\le\frac{1}{3xy+3zx+3\sqrt[3]{xyz}}=\Sigma_{cyc}\frac{1}{3xy+3zx+3}=\Sigma_{cyc}\frac{1}{3\left(xy+zx+1\right)}\)
Dat \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow abc=1\)
\(\Rightarrow A\le\Sigma_{cyc}\frac{1}{3\left(\frac{1}{ab}+\frac{1}{ca}+1\right)}=\Sigma_{cyc}\frac{a}{3\left(a+b+c\right)}=\frac{1}{3}\)
Dau '=' xay ra khi \(x=y=z=1\)
Bạn tham khảo tại đây:
Câu hỏi của Tôi Là Ai - Toán lớp 8 - Học toán với OnlineMath
Ta có:
\(\frac{x}{1+x^2}+\frac{18y}{1+y^2}+\frac{4z}{1+z^2}=xyz\left(\frac{1}{yz\left(1+x^2\right)}+\frac{18}{xz\left(1+y^2\right)}+\frac{4}{xy\left(1+z^2\right)}\right)\)
\(=xyz\left(\frac{1}{yz+x\left(x+y+z\right)}+\frac{18}{xz+y\left(x+y+z\right)}+\frac{4}{xy+z\left(x+y+z\right)}\right)\)
\(=xyz\left(\frac{1}{\left(x+y\right).\left(x+z\right)}+\frac{18}{\left(y+x\right).\left(y+z\right)}+\frac{4}{\left(z+x\right).\left(z+y\right)}\right)\)
\(=xyz.\frac{\left(z+y\right)+18.\left(x+z\right)+4\left(x+y\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)
\(=\frac{xyz\left(22x+5y+19z\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)(đpcm)