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M=\(\left(x_1+x_2\right)^2-2x_1.x_2+\left(y_1+y_2\right)^2-2y_1.y_2\)
Áp dụng định lý viettel :( :v )
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\);\(\left\{{}\begin{matrix}y_1+y_2=-\dfrac{b}{c}\\y_1y_2=\dfrac{a}{c}\end{matrix}\right.\)
\(M=\dfrac{b^2}{a^2}-\dfrac{2c}{a}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}=\dfrac{b^2-4ac}{a^2}+\dfrac{b^2-4ac}{c^2}+2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)
\(\ge2\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge4\)
Dấu = xảy ra: \(\left\{{}\begin{matrix}a=c\\b^2=4ac\end{matrix}\right.\)\(\Leftrightarrow b^2=4a^2=4c^2\)
@_@ 
đưa thẳng câu hỏi luôn đi ; nói như zầy chưa nghỉ ra câu trả lời ; chống mặt chết trước rồi
Bao nhiêu công gõ bài xong rồi đi chơi, chơi về định gửi bài, chơi về bật máy lên gửi thì lỗi, may vãi
Ta có:
\(\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\dfrac{a^2}{2a\left(a+b+c\right)+2a^2+bc}\)
\(\le\dfrac{1}{9}\left(\dfrac{a^2}{a\left(a+b+c\right)}+\dfrac{a^2}{a\left(a+b+c\right)}+\dfrac{a^2}{2a^2+bc}\right)\)
\(=\dfrac{1}{9}\left(\dfrac{2a}{a+b+c}+\dfrac{a^2}{2a^2+bc}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{9}\left(\dfrac{2\left(a+b+c\right)}{a+b+c}+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\)
\(=\dfrac{1}{9}\left(2+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\)
Cần chứng minh \(\dfrac{1}{9}\left(2+\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\right)\le\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ac}+\dfrac{c^2}{2c^2+ab}\le1\)
\(\Leftrightarrow\dfrac{bc}{bc+2a^2}+\dfrac{ca}{ca+2b^2}+\dfrac{ab}{ab+2c^2}\ge1\)
Cauchy-Schwarz: \(VT=\dfrac{bc}{bc+2a^2}+\dfrac{ca}{ca+2b^2}+\dfrac{ab}{ab+2c^2}\)
\(=\dfrac{b^2c^2}{b^2c^2+2a^2bc}+\dfrac{c^2a^2}{c^2a^2+2ab^2c}+\dfrac{a^2b^2}{a^2b^2+2abc^2}\)
\(\ge\dfrac{\left(ab+bc+ca\right)^2}{\left(ab+bc+ca\right)^2}=1\) * Đúng*
Happy New Year (Lunar)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c\left(b+c\right)}{8a^3\left(b+c\right)b^2c}}=\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ca}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2a\left(c+a\right)}{8b^3\left(c+a\right)c^2a}}=\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge3\sqrt[3]{\dfrac{a^2b\left(a+b\right)}{8c^3\left(a+b\right)a^2b}}=\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{1}{4b}+\dfrac{1}{2b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{1}{4c}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{1}{4a}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{3}{4b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{3}{4c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{3}{4a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Rightarrow VT+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
a/ Với mọi số thực ta luôn có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
Lại có do a;b;c là ba cạnh của 1 tam giác nên theo BĐT tam giác ta có:
\(a+b>c\Rightarrow ac+bc>c^2\)
\(a+c>b\Rightarrow ab+bc>b^2\)
\(b+c>a\Rightarrow ab+ac>a^2\)
Cộng vế với vế: \(2\left(ab+bc+ca\right)>a^2+b^2+c^2\)
b/
Do a;b;c là ba cạnh của tam giác nên các nhân tử vế phải đều dương
Ta có:
\(\left(a+b-c\right)\left(b+c-a\right)\le\frac{1}{4}\left(a+b-c+b+c-a\right)^2=b^2\)
Tương tự: \(\left(a+b-c\right)\left(a+c-b\right)\le a^2\)
\(\left(b+c-a\right)\left(a+c-b\right)\le c^2\)
Nhân vế với vế:
\(a^2b^2c^2\ge\left(a+b-c\right)^2\left(b+c-a\right)^2\left(a+c-b\right)^2\)
\(\Leftrightarrow abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right)\)
Bài 1:
Ta có: \(\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}=\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\)
Áp dụng bđt Cauchy Schwarz có:
\(\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+8bc}+b\sqrt{b^2+8bc}+c\sqrt{c^2+8bc}}\)
Lại sử dụng bđt Cauchy schwarz ta có:
\(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}=\sqrt{a}\cdot\sqrt{a^3+8abc}+\sqrt{b}\cdot\sqrt{b^3+8abc}+\sqrt{c}\cdot\sqrt{c^3+8abc}\ge\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}}\)
=> Ta cần chứng minh: \(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)
hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Áp dụng bđt Cosi ta có:
\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)
Nhân các vế của 3 bđt trên ta đc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}=8\sqrt{a^2b^2c^2}=8abc\)
=> Đpcm
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 \(a=b=c=\frac{1}{\sqrt{3}}\)
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