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Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(VT=\dfrac{xz}{y\left(x+z\right)}+\dfrac{xy}{z\left(x+y\right)}+\dfrac{yz}{x\left(y+z\right)}=\dfrac{\left(xz\right)^2}{xyz\left(x+z\right)}+\dfrac{\left(xy\right)^2}{xyz\left(x+y\right)}+\dfrac{\left(yz\right)^2}{xyz\left(y+z\right)}\)
\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=1\)
Cho $a=b=c=1$ thì thỏa mãn đẳng thức nhưng $abc+1=2\neq 0$
Bạn xem lại đề.
a;b;c phải là số dương chứ bạn?
\(\dfrac{a+1}{b^2+1}=a+1-\dfrac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\dfrac{b^2\left(a+1\right)}{2b}=a+1-\dfrac{b+ab}{2}\)
Tương tự:
\(\dfrac{b+1}{c^2+1}\ge b+1-\dfrac{c+bc}{2}\) ; \(\dfrac{c+1}{a^2+1}\ge c+1-\dfrac{a+ca}{2}\)
Cộng vế với vế:
\(VT\ge a+b+c+3-\dfrac{1}{2}\left(a+b+c+ab+bc+ca\right)\)
\(VT\ge6-\dfrac{3}{2}-\dfrac{1}{2}\left(ab+bc+ca\right)\ge\dfrac{9}{2}-\dfrac{1}{6}\left(a+b+c\right)^2=3=a+b+c\)
\(\Rightarrow VT\ge a+b+c\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
\(\dfrac{a^3}{1+b}+\dfrac{1+b}{4}+\dfrac{1}{2}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)}{8\left(a+b\right)}}=\dfrac{3a}{2}\)
\(\dfrac{b^3}{1+c}+\dfrac{1+c}{4}+\dfrac{1}{2}\ge\dfrac{3b}{2}\) ; \(\dfrac{c^3}{1+a}+\dfrac{1+a}{4}+\dfrac{1}{2}\ge\dfrac{3c}{2}\)
\(\Rightarrow VT+\dfrac{a+b+c}{4}+\dfrac{9}{4}\ge\dfrac{3}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{5}{4}\left(a+b+c\right)-\dfrac{9}{4}\ge\dfrac{5}{4}.3\sqrt[3]{abc}-\dfrac{9}{4}=\dfrac{3}{2}\)