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Ta có:\(\sum\dfrac{a^2+6a+3}{a^2+a}=\sum\left(1+\dfrac{5a+3}{a^2+a}\right)=3+\sum\dfrac{5a+3}{a^2+a}\)
Có BĐT phụ: \(\dfrac{5a+3}{a^2+a}\ge-\dfrac{7}{2}a+\dfrac{15}{2}\)đúng vì nó tương đương \(\left(7a+6\right)\left(a-1\right)^2\ge0\left(true\right)\)
Áp dụng tương tự ta có:
\(VT\ge3-\dfrac{7}{2}\left(a+b+c\right)+\dfrac{15}{2}.3\ge3-\dfrac{21}{2}+\dfrac{45}{2}=15\)
Dấu = xảy ra khi a=b=c=1
Ta có : \(9=a^2+a^2+b^2+a^2+b^2+bc+bc+c^2+c^2\ge9\sqrt[9]{a^6\cdot b^6\cdot c^6}=9\sqrt[3]{a^2\cdot b^2\cdot c^2}\Rightarrow abc\le1\) Áp dụng bđt Cô-si vào các số dương : \(a^2+\dfrac{1}{b^2}+\dfrac{1}{b^2}+\dfrac{1}{b^2}\ge4\sqrt[4]{\dfrac{a^2}{b^6}}=4\sqrt{\dfrac{a}{b^3}}\Rightarrow\sqrt{a^2+\dfrac{3}{b^2}}\ge2\cdot\sqrt[4]{\dfrac{a}{b^3}}\)
CM tương tự ta được: \(\sqrt{b^2+\dfrac{3}{c^2}}\ge2\sqrt[4]{\dfrac{b}{c^3}};\sqrt{c^2+\dfrac{3}{a^2}}\ge2\sqrt[4]{\dfrac{c}{a^3}}\Rightarrow P\ge2\cdot\left(\sqrt[4]{\dfrac{a}{b^3}}+\sqrt[4]{\dfrac{b}{c^3}}+\sqrt[4]{\dfrac{c}{a^3}}\right)\ge2\cdot3\cdot\sqrt[12]{\dfrac{a}{b^3}\cdot\dfrac{b}{c^3}\cdot\dfrac{c}{a^3}}=6\sqrt[12]{\dfrac{1}{\left(abc\right)^2}}=6\) Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)
Từ \(a^2b^2+b^2c^2+c^2a^2\ge a^2b^2c^2\)\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=1\)
bài này tui làm rồi ở đây
Bài 1:
Ta có:
\(\text{VT}=\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\)
\(=a-\frac{2ab^2}{a+2b^2}+b-\frac{2bc^2}{b+2c^2}+c-\frac{2ca^2}{c+2a^2}=(a+b+c)-2\left(\frac{ab^2}{a+2b^2}+\frac{bc^2}{b+2c^2}+\frac{ca^2}{c+2a^2}\right)\)
\(=3-2M(*)\)
Áp dụng BĐT Cauchy ta có:
\(M=\frac{ab^2}{a+b^2+b^2}+\frac{bc^2}{b+c^2+c^2}+\frac{ca^2}{c+a^2+a^2}\leq \frac{ab^2}{3\sqrt[3]{ab^4}}+\frac{bc^2}{3\sqrt[3]{bc^4}}+\frac{ca^2}{3\sqrt[3]{ca^4}}\)
\(\Leftrightarrow M\leq \frac{1}{3}(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2})\)
Tiếp tục áp dụng BĐT Cauchy:
\(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2}\leq \frac{ab+ab+1}{3}+\frac{bc+bc+1}{3}+\frac{ca+ca+1}{3}=\frac{2(ab+bc+ac)+3}{3}\)
Mà \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=3\) (quen thuộc)
\(\Rightarrow M\leq \frac{1}{3}.\frac{2.3+3}{3}=1(**)\)
Từ \((*);(**)\Rightarrow \text{VT}\geq 3-2.1=1\)
(đpcm)
Dấu bằng xảy ra khi $a=b=c=1$
Bài 2:
Áp dụng BĐT Cauchy -Schwarz:
\(\text{VT}=\frac{a^3}{a^2+a^2b^2}+\frac{b^3}{b^2+b^2c^2}+\frac{c^3}{c^2+a^2c^2}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{a^2+a^2b^2+b^2+b^2c^2+c^2+c^2a^2}\)
hay:
\(\text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+a^2b^2+b^2c^2+c^2a^2}(*)\)
Mặt khác, theo BĐT Cauchy ta dễ thấy:
\(a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2\)
\(\Rightarrow (a^2+b^2+c^2)^2\geq 3(a^2b^2+b^2c^2+c^2a^2)\)
\(\Leftrightarrow 1\geq 3(a^2b^2+b^2c^2+c^2a^2)\Rightarrow a^2b^2+b^2c^2+c^2a^2\leq \frac{1}{3}(**)\)
Từ \((*);(**)\Rightarrow \text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+\frac{1}{3}}=\frac{3}{4}(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Áp dụng bất đẳng thức Bunyakovsky
\(\Rightarrow\sqrt{\left(\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}\right)\left[\left(\sqrt{2}\right)^2+\left(3\sqrt{2}\right)^2+2^2\right]}\ge\left(\sqrt{\dfrac{4}{a}+9b+ca}\right)^2\)
\(\Leftrightarrow2\sqrt{6}\sqrt{\dfrac{8}{a^2}+\dfrac{9b^2}{2}+\dfrac{c^2a^2}{4}}\ge\dfrac{4}{a}+9b+ac\)
Tương tự ta có \(\left\{{}\begin{matrix}2\sqrt{6}\sqrt{\left(\dfrac{8}{b^2}+\dfrac{9c^2}{2}+\dfrac{a^2b^2}{4}\right)}\ge\dfrac{4}{b}+9c+ab\\2\sqrt{6}\sqrt{\left(\dfrac{8}{c^2}+\dfrac{9a^2}{2}+\dfrac{b^2c^2}{4}\right)}\ge\dfrac{4}{c}+9a+bc\end{matrix}\right.\)
\(\Rightarrow2\sqrt{6}S\ge\dfrac{4}{a}+9a+\dfrac{4}{b}+9b+\dfrac{4}{c}+9c+ab+bc+ac\)
\(\Leftrightarrow2\sqrt{6}S\ge\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{a}+a\ge2\sqrt{4}=4\\\dfrac{4}{b}+b\ge2\sqrt{4}=4\\\dfrac{4}{c}+c\ge2\sqrt{4}=4\end{matrix}\right.\)
\(\Rightarrow\dfrac{4}{a}+a+8a+\dfrac{4}{b}+b+8b+\dfrac{4}{c}+c+8c+ab+bc+ca\ge12+8a+8b+8c+ab+bc+ac\)
\(\Rightarrow2\sqrt{6}S\ge12+8a+8b+8c+ab+bc+ac\)
\(\Leftrightarrow2\sqrt{6}S\ge12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow2a+bc\ge2\sqrt{2abc}\)
Tượng tự ta có \(2b+ac\ge2\sqrt{2abc}\) ; \(2c+ab\ge2\sqrt{2abc}\)
\(\Rightarrow12+2a+bc+2b+ac+2c+ab+6\left(a+b+c\right)\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)
\(\Rightarrow2\sqrt{6}S\ge6\left(a+b+c+\sqrt{2abc}\right)+12\)
Theo đề bài ta có \(a+b+c+\sqrt{2abc}\ge10\)
\(\Rightarrow6\left(a+b+c+\sqrt{2abc}\right)+12\ge72\)
\(\Rightarrow S\ge\dfrac{72}{2\sqrt{6}}=6\sqrt{6}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=2\)
Ta có:
\(\dfrac{a}{bc}+\dfrac{b}{ca}\ge2\sqrt{\dfrac{ab}{abc^2}}=\dfrac{2}{c}\)
Tương tự: \(\dfrac{a}{bc}+\dfrac{c}{ab}\ge\dfrac{2}{b}\) ; \(\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{2}{a}\)
Cộng vế với vế: \(\Rightarrow\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow P\ge\dfrac{a^2+b^2+c^2}{2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a^2+\dfrac{1}{a}+\dfrac{1}{a}\right)+\dfrac{1}{2}\left(a^2+\dfrac{1}{b}+\dfrac{1}{b}\right)+\dfrac{1}{2}\left(c^2+\dfrac{1}{c}+\dfrac{1}{c}\right)\)
\(\Rightarrow P\ge\dfrac{1}{2}.3\sqrt[3]{\dfrac{a^2}{a^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{b^2}{b^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{c^2}{c^2}}=\dfrac{9}{2}\)
\(P_{min}=\dfrac{9}{2}\) khi \(a=b=c=1\)
\(\dfrac{a}{a+2\sqrt{\left(a+bc\right)}}=\dfrac{a}{a+2\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{a}{a+2\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(=\dfrac{a}{a+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\)
\(\le\dfrac{a}{5^2}\left(\dfrac{1}{a}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\right)\)
\(=\dfrac{a}{25}\left(\dfrac{1}{a}+\dfrac{8}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)=\dfrac{1}{25}+\dfrac{8}{25}.\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự:
\(\dfrac{b}{b+2\sqrt{b+ac}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\)
\(\dfrac{c}{c+2\sqrt{c+ab}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(P\le\dfrac{3}{25}+\dfrac{4}{25}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{15}{25}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Bài 1:
a)
\(\sin ^2x+\sin ^2x\cot^2x=\sin ^2x(1+\cot^2x)=\sin ^2x(1+\frac{\cos ^2x}{\sin ^2x})\)
\(=\sin ^2x.\frac{\sin ^2x+\cos^2x}{\sin ^2x}=\sin ^2x+\cos^2x=1\)
b)
\((1-\tan ^2x)\cot^2x+1-\cot^2x\)
\(=\cot^2x(1-\tan^2x-1)+1=\cot^2x(-\tan ^2x)+1=-(\tan x\cot x)^2+1\)
\(=-1^2+1=0\)
c)
\(\sin ^2x\tan x+\cos^2x\cot x+2\sin x\cos x=\sin ^2x.\frac{\sin x}{\cos x}+\cos ^2x.\frac{\cos x}{\sin x}+2\sin x\cos x\)
\(=\frac{\sin ^3x}{\cos x}+\frac{\cos ^3x}{\sin x}+2\sin x\cos x=\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin x\cos x}=\frac{(\sin ^2x+\cos ^2x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}\)
\(=\frac{1}{\frac{\sin 2x}{2}}=\frac{2}{\sin 2x}\)
Bài 2:
Áp dụng BĐT Cauchy Schwarz ta có:
\(P=\frac{a^2}{\sqrt{a(2c+a+b)}}+\frac{b^2}{\sqrt{b(2a+b+c)}}+\frac{c^2}{\sqrt{c(2b+c+a)}}\)
\(\geq \frac{(a+b+c)^2}{\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}}(*)\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\((\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq (a+b+c)(2c+a+b+2a+b+c+2b+c+a)\)
\(\Leftrightarrow (\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq 4(a+b+c)^2\)
\(\Rightarrow \sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}\leq 2(a+b+c)(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)
Dấu "=" xảy ra khi $a=b=c=1$