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\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{1-a-b-ab}}=\sqrt{\dfrac{ab}{\left(1-b\right)\left(1-a\right)}}\le\dfrac{\dfrac{a}{1-b}+\dfrac{b}{1-a}}{2}\left(1\right)\) \(tương-tự\Rightarrow\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{\dfrac{b}{1-c}+\dfrac{c}{1-b}}{2}\left(2\right)\)
\(\Rightarrow\sqrt{\dfrac{ca}{b+ ca}}\le\dfrac{\dfrac{c}{1-a}+\dfrac{a}{1-c}}{2}\left(3\right)\)
\( \left(1\right)\left(2\right)\left(3\right)\Rightarrow A\le\dfrac{\dfrac{a}{1-b}+\dfrac{b}{1-a}+\dfrac{b}{1-c}+\dfrac{c}{1-b}+\dfrac{c}{1-a}+\dfrac{a}{1-c}}{2}=\dfrac{\dfrac{a+c}{1-b}+\dfrac{b+c}{1-a}+\dfrac{b+a}{1-c}}{2}=\dfrac{\dfrac{1-b}{1-b}+\dfrac{1-a}{1-a}+\dfrac{1-c}{1-c}}{2}=\dfrac{3}{2}\)
\(\Rightarrow A_{max}=\dfrac{3}{2}\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Em tham khảo ở đây:
Cho a,b,c > 0 và ab + bc + ac = 1. Chứng minh rằng :\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^... - Hoc24
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(P=\frac{a\sqrt{a}}{\sqrt{2c+a+b}}+\frac{b\sqrt{b}}{\sqrt{2a+b+c}}+\frac{c\sqrt{c}}{\sqrt{2b+c+a}}\)
\(P=\frac{a^2}{\sqrt{a(2c+a+b)}}+\frac{b^2}{\sqrt{b(2a+b+c)}}+\frac{c^2}{\sqrt{c(2b+c+a)}}\)
\(\geq \frac{(a+b+c)^2}{\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+a+c)}}\)
Áp dụng BĐT Bunhiacopxky:
\((\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+a+c)})^2\leq (a+b+c[((2c+a+b)+(2a+b+c)+(2b+a+c)]\)
\(\Leftrightarrow (\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+a+c)})^2\leq 4(a+b+c)^2\)
\(\Leftrightarrow \sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+a+c)}\leq 2(a+b+c)\)
Do đó:
\(P\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)
Dấu bằng 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 \(\Leftrightarrow a=b=c=1\)
Bài 1:
Áp dụng BĐT Bunhiacopxky ta có:
\((a^2+2c^2)(1+2)\geq (a+2c)^2\)
\(\Rightarrow \sqrt{a^2+2c^2}\geq \frac{a+2c}{\sqrt{3}}\)
\(\Rightarrow \frac{\sqrt{a^2+2c^2}}{ac}\geq \frac{a+2c}{\sqrt{3}ac}=\frac{ab+2bc}{\sqrt{3}abc}\)
Hoàn toàn tương tự: \(\left\{\begin{matrix} \frac{\sqrt{c^2+2b^2}}{bc}\geq \frac{ac+2ab}{\sqrt{3}abc}\\ \frac{\sqrt{b^2+2a^2}}{ab}\geq \frac{bc+2ac}{\sqrt{3}abc}\end{matrix}\right.\)
Cộng theo vế các BĐT trên thu được:
\(\text{VT}\geq \frac{1}{\sqrt{3}}.\frac{ab+2bc+ac+2ab+bc+2ac}{abc}=\frac{1}{\sqrt{3}}.\frac{3(ab+bc+ac)}{abc}=\frac{1}{\sqrt{3}}.\frac{3abc}{abc}=\sqrt{3}\)
Ta có đpcm
Dấu bằng xảy ra khi $a=b=c=3$
Bài 2: Bài này sử dụng pp xác định điểm rơi thôi.
Áp dụng BĐT AM-GM ta có:
\(24a^2+24.(\frac{31}{261})^2\geq 2\sqrt{24^2.(\frac{31}{261})^2a^2}=\frac{496}{87}a\)
\(b^2+(\frac{248}{87})^2\geq 2\sqrt{(\frac{248}{87})^2.b^2}=\frac{496}{87}b\)
\(93c^2+93.(\frac{8}{261})^2\geq 2\sqrt{93^2.(\frac{8}{261})^2c^2}=\frac{496}{87}c\)
Cộng theo vế:
\(B+\frac{248}{29}\geq \frac{496}{87}(a+b+c)=\frac{496}{87}.3=\frac{496}{29}\)
\(\Rightarrow B\geq \frac{496}{29}-\frac{248}{29}=\frac{248}{29}\)
Vậy \(B_{\min}=\frac{248}{29}\). Dấu bằng xảy ra khi: \((a,b,c)=(\frac{31}{261}; \frac{248}{87}; \frac{8}{261})\)
\(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\)
\(=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\dfrac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\dfrac{ca}{b\left(a+b+c\right)+ca}}\)
\(=\sqrt{\dfrac{ab}{\left(b+c\right)\left(c+a\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{a+b}\right)=\dfrac{1}{2}\)
\("=" \Leftrightarrow a=b=c=\frac{1}{3}\)
Câu hỏi của lê thị tiều thư - Toán lớp 9 | Học trực tuyến
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 $a=b=c=1$
\(A=\dfrac{a+b+c}{a+\sqrt{\dfrac{a}{2}.2b}+\sqrt[3]{\dfrac{a}{4}.b.4c}}\ge\dfrac{a+b+c}{a+\dfrac{1}{2}\left(\dfrac{a}{2}+2b\right)+\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)}=\dfrac{3}{4}\)
thầy cho em hỏi làm sao có thể tách căn ab với 3 căn abc ra được như vậy , ý em là làm sao để chọn được số 4 nhân vào ạ