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Bài 1: Theo đề bài: \(VT=\left(a-1\right)+\frac{1}{\left(a-1\right)}+1\ge2\sqrt{\left(a-1\right).\frac{1}{a-1}}+1=2+1=3^{\left(đpcm\right)}\)
Đẳng thức xảy ra khi \(\left(a-1\right)=\frac{1}{a-1}\Leftrightarrow a=2\)
Bài 2: \(BĐT\Leftrightarrow\left(a^2+2\right)^2\ge4\left(a^2+1\right)\)
\(\Leftrightarrow a^4+4a^2+4\ge4a^2+4\)
\(\Leftrightarrow a^4\ge0\) (đúng). Đẳng thức xảy ra khi a = 0
Bài 3: Hình như sai đề thì phải ạ. Nếu a = 1,5 ; b = 1 thì \(\frac{19}{10}=1,9< 3\)
\(\left\{{}\begin{matrix}a>0\\\frac{a}{b}>1\end{matrix}\right.\) \(\Rightarrow b>0\Rightarrow a>b\Rightarrow a-b>0\)
\(\Rightarrow4.b\left(a-b\right)\le\left(b+a-b\right)^2=a^2\)
\(\Rightarrow P=\frac{2a^3+1}{4b\left(a-b\right)}\ge\frac{2a^3+1}{a^2}=2a+\frac{1}{a^2}=a+a+\frac{1}{a^2}\ge3\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=1\\b=\frac{1}{2}\end{matrix}\right.\)
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\(VT=a+\frac{4}{\left(a-b\right)\left(b+1\right)^2}\)
\(=a-b+\frac{4}{\left(a-b\right)\left(b+1\right)^2}+\frac{b+1}{2}+\frac{b+1}{2}-1\)
\(\ge4\sqrt[4]{a-b\cdot\frac{4}{\left(a-b\right)\left(b+1\right)^2}\cdot\frac{b+1}{2}\cdot\frac{b+1}{2}}-1\)
\(\ge4-1=3=VP\)
Áp dụng BĐT Svarxơ:
\(\Sigma\frac{a^2}{\sqrt{5-2\left(b+c\right)}}\ge\frac{\left(a+b+c\right)^2}{\sqrt{5-2\left(b+c\right)}+\sqrt{5-2\left(a+c\right)}+\sqrt{5-2\left(a+b\right)}}\)\(\frac{3^2}{\sqrt{5-2\left(b+c\right)}+\sqrt{5-2\left(a+c\right)}+\sqrt{5-2\left(b+c\right)}}\)
Có: \(\sqrt{5-2\left(b+c\right)}=\sqrt{2\left(1-\left(3-a\right)\right)+3}\)\(=\sqrt{-4+2a+3}=\sqrt{2a-1}\)
CMTT: \(\sqrt{5-2\left(a+c\right)}=\sqrt{2b-1}\);\(\sqrt{5-2\left(a+b\right)}=\sqrt{2c-1}\)
\(\Rightarrow\Sigma\frac{a^2}{\sqrt{5-2\left(b+c\right)}}\ge\frac{9}{\sqrt{2a-1}+\sqrt{2b-1}+\sqrt{2c-1}}\)\(\ge\frac{9}{\sqrt{\left(1^2+1^2+1^2\right)\left(2a-1+2b-1+2c-1\right)}}\)(BDT Bunhiacopxki)\(=\frac{9}{\sqrt{3\left[2\left(a+b+c\right)-3\right]}}=\frac{9}{\sqrt{3\left[6-3\right]}}=\frac{9}{3}=3\)(dpcm)
Áp dụng BĐT AM-GM ta có:
\(VT=a^2+b^2+\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b\)
\(=1+\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b\)
\(=1+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{1}{a}+2a\right)+\left(\frac{1}{b}+2b\right)-\left(a+b\right)\)
\(\ge3+2\sqrt{\frac{1}{a}\cdot2a}+2\sqrt{\frac{1}{b}\cdot2b}-\sqrt{2\left(a^2+b^2\right)}\)
\(\ge3+4\sqrt{2}-\sqrt{2}=3+3\sqrt{2}=3\left(1+\sqrt{2}\right)\)
Khi \(a=b=\frac{1}{\sqrt{2}}\)
Áp dụng BĐT Cô-si cho 3 số dương ta có:
\(\left(1+\frac{1}{a}\right)^4+\left(1+\frac{1}{b}\right)^4+\left(1+\frac{1}{c}\right)^4\ge3\left(\sqrt[3]{\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)}\right)^4\)
Ta chứng minh: \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\ge\left(1+\frac{3}{2+abc}\right)^3\left(1\right)\)
Theo BĐT Cô - si ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\)
\(\ge1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{\left(abc\right)^2}}+\frac{1}{abc}=\left(1+\frac{1}{\sqrt[3]{abc}}\right)^3\ge\left(1+\frac{3}{2+abc}\right)^3\)
(Vì \(abc+2=abc+1+1\ge3\sqrt[3]{abc}\))
Vậy \(\left(1\right)\) được chứng minh \(\Rightarrow BĐT\) đúng \(\forall a,b,c>0\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=c=1\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow VT\ge3\sqrt[3]{\left[\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\right]^4}\)
\(\Rightarrow VT\ge3\left(\sqrt[3]{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}}\right)^4\left(1\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\\\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3\sqrt[3]{\frac{1}{a^2b^2c^2}}\end{cases}}\)
\(\Rightarrow1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\ge1+3\sqrt[3]{\frac{1}{abc}}\)
\(+3\sqrt[3]{\frac{1}{a^2b^2c^2}}+\frac{1}{abc}\)
\(\Rightarrow1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}\ge\left(1+\frac{1}{\sqrt[3]{abc}}\right)^3\)
\(\Rightarrow3\left(\sqrt[3]{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc}}\right)^4\)
\(\ge3\left(1+\frac{1}{\sqrt[3]{abc}}\right)^4\)
\(\left(2\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\sqrt[3]{abc}\le\frac{abc+1+1}{3}=\frac{abc+2}{3}\)
\(\Rightarrow1+\frac{1}{\sqrt[3]{abc}}\ge1+\frac{3}{abc+2}\)
\(\Rightarrow3\left(1+\frac{1}{\sqrt[3]{abc}}\right)^4\ge3\left(1+\frac{3}{abc+2}\right)^4\left(3\right)\)
Từ (1) , (2) và (3)
\(\Rightarrow VT\ge3\left(1+\frac{3}{abc+2}\right)^4\)
\(\Leftrightarrow\left(1+\frac{1}{a}\right)^4+\left(1+\frac{1}{b}\right)^4+\left(1+\frac{1}{c}\right)^4\ge3\left(1+\frac{3}{2+abc}\right)^4\left(đpcm\right)\)
Chúc bạn học tốt !!!
Ta có:
\(21b+\frac{3}{a}=\frac{3}{a}+\frac{a}{3}+\frac{62a}{3}\ge2\sqrt{\frac{3}{a}.\frac{a}{3}}+\frac{62.3}{3}=2+62=64\left(a\ge3\right)\left(1\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{3}{a}=\frac{a}{3}\)và \(a=3\Leftrightarrow a=3\)
\(\frac{21}{b}+3b=\frac{21}{b}+\frac{7b}{3}+\frac{2b}{3}\ge2\sqrt{\frac{21}{b}.\frac{7b}{3}}+\frac{2.3}{3}=14+2=16\left(b\ge3\right)\left(2\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{21}{b}=\frac{7b}{3}\)và \(b=3\Leftrightarrow b=3\)
Từ (1) và (2) suy ra điều cần chứng minh.
Dấu "=" xảy ra \(\Leftrightarrow a=b=3\)