Question

cho a, b, c là các số không âm. Chứng minh rằng:

\(\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le1\)

Answer 1

Ta chứng minh bất đẳng thức: \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)  (a,b,c,x,y,z dương)    (Hệ quả của bất đẳng thức Cauchy-Schwarz (Bunyakovsky))

\(\left[\frac{a^2}{\left(\sqrt{x}\right)^2}+\frac{b^2}{\left(\sqrt{y}\right)^2}+\frac{c^2}{\left(\sqrt{z}\right)^2}\right]\left[\left(\sqrt{x}\right)^2+\sqrt{y}^2+\sqrt{z^2}\right]\ge a^2+b^2+c^2\)

\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)

Ta có:

\(A=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ac}+\frac{ab}{c^2+2ab}\)

\(2A=\frac{2bc}{a^2+2bc}+\frac{2ca}{b^2+2ac}+\frac{2ab}{c^2+2ab}\)

\(=\frac{a^2+2bc-a^2}{a^2+2bc}+\frac{b^2+2ca-b^2}{b^2+2ac}+\frac{c^2+2ab-c^2}{c^2+2ab}\)

\(=3-\left(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\right)\)

\(\le3-\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2ac+2bc}=3-\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=3-1=2\)

=> A<=1 

Answer 2

a,b,c dương 

Ta viết lại BĐT thành: \(\frac{1}{\frac{a^2}{bc}+2}+\frac{1}{\frac{b^2}{ca}+2}+\frac{1}{\frac{c^2}{ab}+2}\le1\)

Đặt \(\frac{a^2}{bc}=x;\frac{b^2}{ca}=y;\frac{c^2}{ab}=z\Rightarrow\hept{\begin{cases}x,y,z>0\\xyz=1\end{cases}}\)và ta cần chứng minh \(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\le1\)

Xét biểu thức\(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\) \(\frac{\left(y+2\right)\left(z+2\right)+\left(z+2\right)\left(x+2\right)+\left(x+2\right)\left(y+2\right)}{\left(x+2\right)\left(y+2\right)\left(z+2\right)}\)

\(=\frac{\left(yz+2y+2z+4\right)+\left(zx+2z+2x+4\right)+\left(xy+2x+2y+4\right)}{\left(xy+2x+2y+4\right)\left(z+2\right)}\)

\(=\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{xyz+2\left(xy+yz+zx\right)+4\left(x+y+z\right)+8}\)\(=\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{xyz+\left(xy+yz+zx\right)+\left(xy+yz+zx\right)+4\left(x+y+z\right)+8}\)\(\le\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{xyz+3\sqrt{\left(xyz\right)^2}+\left(xy+yz+zx\right)+4\left(x+y+z\right)+8}\)\(=\frac{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}{\left(xy+yz+zx\right)+4\left(x+y+z\right)+12}=1\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c

Answer 3

Ta viết lại BĐT thành: \(\frac{1}{\frac{a^2}{bc}+2}+\frac{1}{\frac{b^2}{ca}+2}+\frac{1}{\frac{c^2}{ab}+2}\le1\)

Đặt \(\frac{a^2}{bc}=x^2;\frac{b^2}{ca}=y^2;\frac{c^2}{ab}=z^2\)thì \(xyz=1\)

Khi đó BĐT chuyển thành dạng:\(\frac{1}{x^2+2}+\frac{1}{y^2+2}+\frac{1}{z^2+2}\le1\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x^2+2}+\frac{1}{2}-\frac{1}{y^2+2}+\frac{1}{2}-\frac{1}{z^2+2}\ge\frac{3}{2}-1=\frac{1}{2}\)

\(\Leftrightarrow\frac{x^2}{x^2+2}+\frac{y^2}{y^2+2}+\frac{z^2}{z^2+2}\ge1\)

Áp dụng BĐT Bunyakovsky dạng phân thức, ta được: \(\frac{x^2}{x^2+2}+\frac{y^2}{y^2+2}+\frac{z^2}{z^2+2}\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+6}\)

Đến đây, ta cần chỉ ra rằng \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+6}\ge1\Leftrightarrow xy+yz+zx\ge3\)(Đúng theo BĐT AM - GM vì \(xy+yz+zx\ge3\sqrt[3]{\left(xyz\right)^2}=3\)

Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c