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Từ giả thiết, ta có:
\(\dfrac{1}{1+a}\ge1-\dfrac{1}{1+b}+1-\dfrac{1}{1+c}+1-\dfrac{1}{1+d}=\dfrac{b}{1+b}+\dfrac{c}{c+1}+\dfrac{d}{d+1}\ge3\sqrt[3]{\dfrac{b.c.d}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)
Tương tự:
\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{cda}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)
\(\dfrac{1}{1+c}\ge3\sqrt[3]{\dfrac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\dfrac{1}{1+d}\ge3\sqrt[3]{\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Nhân vế theo vế 4 BĐT vừa chứng minh rồi rút gọn ta được:
\(abcd\le\dfrac{1}{81}\left(đpcm\right)\)
\(\dfrac{b}{1+b}+\dfrac{c}{1+c}+\dfrac{d}{1+d}\le1-\dfrac{a}{1+a}=\dfrac{1}{1+a}\)
\(\Rightarrow\dfrac{1}{1+a}\ge\dfrac{b}{1+b}+\dfrac{c}{1+c}+\dfrac{d}{1+d}\ge3\dfrac{\sqrt[3]{bcd}}{\sqrt[3]{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)
Chứng minh tương tự ta có:
\(\dfrac{1}{1+b}\ge3\dfrac{\sqrt[3]{acd}}{\sqrt[3]{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)
\(\dfrac{1}{1+c}\ge3\dfrac{\sqrt[3]{abd}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\dfrac{1}{1+d}\ge3\dfrac{\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Nhân vế với vế của các BĐT trên ta được:
\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge81\dfrac{abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\)
\(\Rightarrow81abcd\le1\Rightarrow abcd\le\dfrac{1}{81}\)
Dấu "=" xảy ra khi \(a=b=c=d=\dfrac{1}{3}\)
Ta có BĐT phụ \(\dfrac{1+\sqrt{a}}{1-a}\ge4a+1\)
\(\Leftrightarrow-\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)^2}{\sqrt{a}-1}\ge0\forall\dfrac{1}{4}< a< 0\)
Tương tự cho 3 BĐT còn lại ta cũng có:
\(\dfrac{1+\sqrt{b}}{1-b}\ge4b+1;\dfrac{1+\sqrt{c}}{1-c}\ge4c+1;\dfrac{1+\sqrt{d}}{1-d}\ge4d+1\)
Cộng theo vế 4 BĐT trên ta có:
\(VT\ge4\left(a+b+c+d\right)+4=8=VP\)
Xảy ra khi \(a=b=c=d=\dfrac{1}{4}\)
Ta cần chứng minh :
\(\dfrac{1+\sqrt{a}}{1-a}\ge4a+1\) \(\forall a\in\left(0;\dfrac{1}{4}\right)\)
\(\Leftrightarrow1+\sqrt{a}\ge\left(4a+1\right)\left(1-a\right)\)
\(\Leftrightarrow1+\sqrt{a}\ge4a-4a^2+1-a\)
\(\Leftrightarrow4a^2-4a-1+a+1+\sqrt{a}\ge0\)
\(\Leftrightarrow4a^2-3a+\sqrt{a}\ge0\)
\(\Leftrightarrow\left(4a^2-a\right)-\left(2a-\sqrt{a}\right)\ge0\)
\(\Leftrightarrow\left(2a-\sqrt{a}\right)\left(2a+\sqrt{a}\right)-\left(2a-\sqrt{a}\right)\ge0\)
\(\Leftrightarrow\left(2a-\sqrt{a}\right)\left(2a+\sqrt{a}-1\right)\ge0\)
Ta có: \(2a-\sqrt{a}=\left(\sqrt{2a}-\dfrac{\sqrt{2}}{4}\right)^2-\dfrac{1}{8}\ge0\) \(\forall a\in\left(0;\dfrac{1}{4}\right)\)
\(\left(2a+\sqrt{a}-1\right)=\left(\sqrt{2a}+\dfrac{\sqrt{2}}{4}\right)^2-\dfrac{9}{8}\ge0\)
\(\forall a\in\left(0;\dfrac{1}{4}\right)\)
Vậy: \(\dfrac{1+\sqrt{a}}{1-a}\ge4a+1\) \(\forall a\in\left(0;\dfrac{1}{4}\right)\)
Tương tự: \(\dfrac{1+\sqrt{b}}{1-b}\ge4b+1\forall b\in\left(0;1\right)\)
\(\dfrac{1+\sqrt{c}}{1-c}\ge4c+1\forall c\in\left(0;\dfrac{1}{4}\right)\)
\(\dfrac{1+\sqrt{d}}{1-d}\ge4d+1\forall d\in\left(0;\dfrac{1}{4}\right)\)
Cộng các BĐT vừa chứng minh, ta được:
\(\dfrac{1+\sqrt{a}}{1-a}+\dfrac{1+\sqrt{b}}{1-b}+\dfrac{1+\sqrt{c}}{1-c}+\dfrac{1+\sqrt{d}}{1-d}\ge4\left(a+b+c+d\right)+4=8\)
Vậy: Ta suy ra được điều phải chứng minh
Bài 1:
\(P=(x+1)\left(1+\frac{1}{y}\right)+(y+1)\left(1+\frac{1}{x}\right)\)
\(=2+x+y+\frac{x}{y}+\frac{y}{x}+\frac{1}{x}+\frac{1}{y}\)
Áp dụng BĐT Cô-si:
\(\frac{x}{y}+\frac{y}{x}\geq 2\)
\(x+\frac{1}{2x}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
\(y+\frac{1}{2y}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
Áp dụng BĐT SVac-xơ kết hợp với Cô-si:
\(\frac{1}{2x}+\frac{1}{2y}\geq \frac{4}{2x+2y}=\frac{2}{x+y}\geq \frac{2}{\sqrt{2(x^2+y^2)}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Cộng các BĐT trên :
\(\Rightarrow P\geq 2+2+\sqrt{2}+\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
Vậy \(P_{\min}=4+3\sqrt{2}\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)
Bài 2:
Áp dụng BĐT Svac-xơ:
\(\frac{1}{a+3b}+\frac{1}{b+a+2c}\geq \frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)
\(\frac{1}{b+3c}+\frac{1}{b+c+2a}\geq \frac{4}{2b+4c+2a}=\frac{2}{b+2c+a}\)
\(\frac{1}{c+3a}+\frac{1}{c+a+2b}\geq \frac{4}{2c+4a+2b}=\frac{2}{c+2a+b}\)
Cộng theo vế và rút gọn :
\(\Rightarrow \frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\geq \frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
\(\dfrac{25112012}{11}=2282911-\dfrac{9}{11}\Rightarrow a=2282911\)
\(\dfrac{9}{11}=\dfrac{1}{\dfrac{11}{9}}=\dfrac{1}{1+\dfrac{2}{9}}=\dfrac{1}{1+\dfrac{1}{\dfrac{9}{2}}}=\dfrac{1}{1+\dfrac{1}{4+\dfrac{1}{2}}}\)
\(\Rightarrow b=1;c=4;d=2\)