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\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có:
\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2\sqrt{\dfrac{ab}{c}.\dfrac{bc}{a}}=2b\)
Tương tự: \(\dfrac{ab}{c}+\dfrac{ca}{b}\ge2a\) ; \(\dfrac{bc}{a}+\dfrac{ca}{b}\ge2c\)
Cộng vế:
\(2P\ge2\left(a+b+c\right)\Rightarrow P\ge a+b+c=1\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
Tương tự: \(\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\) ; \(\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)\)
Cộng vế với vế:
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Cho a, b, c, d là các chữ số thỏa mãn: ab+ca=da ab-ca=a Tìm giá trị của d.
\(không\) \(dùng\) \(bđt\) \(làm\) \(sao\) \(ra\) \(được\) ??
\(\sqrt{a^2+\dfrac{1}{b^2}}=\dfrac{1}{\sqrt{17}}.\sqrt{\left(1+4^2\right)\left(a^2+\dfrac{1}{b^2}\right)}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\left(bunhiacopki\right)\)
\(tương-tự:\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\)
\(\sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\)
\(\Rightarrow Q\ge\dfrac{1}{\sqrt{17}}\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{\sqrt{17}}\left[16a+\dfrac{4}{a}+16b+\dfrac{4}{b}+16c+\dfrac{4}{c}-15\left(a+b+c\right)\right]\)
\(bđt:cosi\Rightarrow16a+\dfrac{4}{a}\ge2\sqrt{16a.\dfrac{4}{a}}=2\sqrt{16.4}=16\)
\(tương-tự\Rightarrow16b+\dfrac{4}{b}\ge16;16c+\dfrac{4}{c}\ge16\)
\(có:a+b+c\le\dfrac{3}{2}\Rightarrow15\left(a+b+c\right)\le\dfrac{45}{2}\)
\(\Rightarrow-15\left(a+b+c\right)\ge-\dfrac{45}{2}\)
\(\Rightarrow Q\ge\dfrac{1}{\sqrt{17}}\left(16+16+16-\dfrac{45}{2}\right)=\dfrac{3\sqrt{17}}{2}\)
\(dấu"="xayra\Leftrightarrow a=b=c=\dfrac{1}{2}\)
các bước ban đầu dùng bunhia chọn được 1+4^2 là do dự đoán được trước điểm rơi tại a=b=c=1/2 thôi bạn,cả bước tách dùng cosi cũng dự đoán dc điểm rơi =1/2 nên tách đc thôi
Tại sao lại k được dùng nhỉ? Trông khi dùng thì bài toán sẽ dễ giải quyết hơn
Áp dụng Bunhiacopxki:
\(\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(\dfrac{1}{4}+4\right)}\ge\dfrac{a}{2}+\dfrac{2}{b}\)
\(\Rightarrow\sqrt{a^2+\dfrac{1}{b^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{a}{2}+\dfrac{2}{b}\right)\)
Do đó:
\(Q\ge\dfrac{2}{\sqrt{17}}\left[\dfrac{a+b+c}{2}+2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)
\(\Rightarrow Q\ge\dfrac{2}{\sqrt{17}}\left[\dfrac{a+b+c}{2}+\dfrac{18}{a+b+c}\right]\)
Áp dụng Cô-si:
\(\dfrac{a+b+c}{2}+\dfrac{9}{8\left(a+b+c\right)}\ge\dfrac{3}{2}\)
Do đó:
\(Q\ge\dfrac{2}{\sqrt{17}}\left[\dfrac{3}{2}+\dfrac{135}{8\left(a+b+c\right)}\right]\ge\dfrac{2}{\sqrt{17}}\left[\dfrac{3}{2}+\dfrac{135}{8.\dfrac{3}{2}}\right]=\dfrac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Thỏa mãn $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ hay $a+b+c=1$ vậy bạn?
\(P=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\) (BĐT Cauchy Schwarz)
\(=\dfrac{9}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{7}{ab+bc+ca}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2ac+2bc}+\dfrac{7}{ab+bc+ca}\)
\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)
Ta có: \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\) .Thế vào biểu thức
\(\Rightarrow P\ge9+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)
\(\Rightarrow P_{min}=30\) khi \(a=b=c=\dfrac{1}{3}\)
mik cảm ơn