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a/ \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ; \(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c}\) ; \(\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)
Cộng theo vế :
\(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
b/ \(\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{4}{a+2b+c}\)
\(\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{4}{b+2c+a}\)
\(\frac{1}{c+a}+\frac{1}{a+b}\ge\frac{4}{c+b+2a}\)
Cộng theo vế :
\(2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge4\left(\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\right)\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge2\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\)
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
a/ \(VT=\frac{1}{a+a+b+c}+\frac{1}{a+b+b+c}+\frac{1}{a+b+c+c}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow VT\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{3}{4}\)
b/ \(VT\le\frac{ab}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{bc}{4}\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{ca}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(VT\le\frac{a}{4}+\frac{b}{4}+\frac{b}{4}+\frac{c}{4}+\frac{c}{4}+\frac{a}{4}=\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
1)
\(2a+\frac{4}{a}+\frac{16}{a+2}=\left(a+\frac{4}{a}\right)+\left[\left(a+2\right)+\frac{16}{a+2}\right]-2\ge4+8-2=10\)
Dấu "=" xảy ra khi a=2
2)
\(\hept{\begin{cases}\sqrt{a\left(1-4a\right)}=\frac{1}{2}\sqrt{4a\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4a+1-4a}{2}=\frac{1}{4}\\\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\\\sqrt{c\left(1-4c\right)}=\frac{1}{2}\sqrt{4c\left(1-4c\right)}\le\frac{1}{2}\cdot\frac{4c+1-4c}{2}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{8}\)
Mấy cái dấu "=" anh tự xét.
Áp dụng BĐT AM-GM: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)
a) Áp dụng: \(VT\ge\frac{\left(a+b+c\right)^2}{3}.\frac{9}{2\left(a+b+c\right)}=\frac{3}{2}\left(a+b+c\right)\)
b) \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{x+y+z+3}=\frac{3}{4}\)
\(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\) ( đúng )
Dấu "=" \(\Leftrightarrow a=b\)
a) Áp dụng BĐT trên ta có:
\(\Sigma\left(\frac{1}{a^3+b^3+abc}\right)\le\Sigma\left(\frac{1}{ab\left(a+b\right)+abc}\right)=\Sigma\left[\frac{1}{ab}\cdot\left(\frac{1}{a+b+c}\right)\right]=\frac{1}{a+b+c}\cdot\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{a+b+c}{\left(a+b+c\right)\cdot abc}=\frac{1}{abc}\)
Dấu "=" khi \(a=b=c\)
b) \(\Sigma\left(\frac{1}{a^3+b^3+1}\right)\le\Sigma\left(\frac{1}{ab\left(a+b\right)+abc}\right)=\Sigma\left[\frac{1}{ab}\cdot\left(\frac{1}{a+b+c}\right)\right]=\frac{1}{abc}=1\)
Dấu "=" khi \(a=b=c=1\)
c) \(\Sigma\left(\frac{1}{a+b+1}\right)\le\Sigma\left(\frac{1}{\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}\right)+\sqrt[3]{abc}}\right)=\Sigma\left[\frac{1}{\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)}\right]\)
\(=\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}\cdot\left(\frac{1}{\sqrt[3]{ab}}+\frac{1}{\sqrt[3]{bc}}+\frac{1}{\sqrt[3]{ca}}\right)=\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\cdot\sqrt[3]{abc}}=\frac{1}{\sqrt[3]{abc}}=1\)
Dấu "=" khi \(a=b=c=1\)
1.
C/m bổ đề: \(a^3-b^3\ge\frac{1}{4}\left(a^3-b^3\right)\) với \(\forall a,b\in R,a\ge b\)
\(\Leftrightarrow4a^3-4b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\ge0\)
\(\Leftrightarrow3a^3+3a^2b-3ab^2-3b^3\ge0\)
\(\Leftrightarrow3\left(a^2-b^2\right)\left(a+b\right)\ge0\)
\(\Leftrightarrow3\left(a+b\right)^2\left(a-b\right)\ge0\)(đúng)
Theo bài ra: \(a^3-b^3\ge3a-3b-4\)
\(\Leftrightarrow\) Cần c/m: \(\left(a-b\right)^3\ge12a-12b-16\)(1)
Thật vậy:
\(\left(1\right)\)\(\Leftrightarrow\left(a-b\right)^3-12\left(a-b\right)+16\ge0\)
\(\Leftrightarrow\left[\left(a-b\right)^3-8\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a-b\right)+4\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a+b\right)-8\right]\ge0\)
\(\Leftrightarrow\left(a-b-2\right)^2\left(a-b+4\right)\ge0\) (đúng với mọi a,b thỏa mãn \(a,b\in R,a\ge b\))
2.
\(BĐT\Leftrightarrow\frac{1}{\frac{a+b}{ab}}+\frac{1}{\frac{c+d}{cd}}\le\frac{1}{\frac{a+b+c+d}{\left(a+c\right)\left(b+d\right)}}\)
\(\Leftrightarrow\frac{ab}{a+b}+\frac{cd}{c+d}\le\frac{\left(a+c\right)\left(b+d\right)}{a+b+c+d}\)
\(\Leftrightarrow\frac{ab\left(c+d\right)+cd\left(a+b\right)}{\left(a+b\right)\left(c+d\right)}\le\)\(\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\frac{abc+abd+acd+bcd}{ac+ad+bc+bd}\le\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\left(ad+ab+bc+cd\right)\left(ac+ad+bc+bd\right)\ge\)\(\left(a+b+c+d\right)\left(abc+abd+acd+bcd\right)\)
\(\Leftrightarrow\left(ad\right)^2-2abcd+\left(bc\right)^2\ge0\)
\(\Leftrightarrow\left(ad-bc\right)^2\ge0\) (đúng với mọi a,b,c,d>0)
a) Áp dụng BĐT Cauchy-Schwarz dạng Engel: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Tương tự:\(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c};\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)
Cộng theo vế 3 BĐT trên rồi chia cho 2 ta thu được đpcm
Đẳng thức xảy ra khi \(a=b=c\)
b)Đặt \(a+b=x;b+c=y;c+a=z\). Cần chứng minh:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Cách làm tương tự câu a.
c) \(VT=\Sigma_{cyc}\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\Sigma_{cyc}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\le\frac{1}{16}\Sigma\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
Đẳng thức xảy ra khi \(a=b=c=\frac{3}{4}\)
d) Em làm biếng quá anh làm nốt đi:P
lm phần d đi a k bt lm