\(=\left(1^2+4^2\right)\left(a^2+\dfrac{1}{b^2}\right)\ge\left(1a+4.\dfrac{1}{b}\right)^2\\ \Rightarrow\sqrt{a^2+\dfrac{1}{vb^2}}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\) 

Tương tự

\(\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\\ \sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\\ Do.đó:\\ Q\ge\dfrac{1}{\sqrt{17}}\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)\ge\dfrac{1}{\sqrt{17}}\\ \left(a+b+c+\dfrac{36}{a+b+c}\right)\) 

\(=\dfrac{1}{\sqrt{17}}\\ \left[a+b+c+\dfrac{9}{4\left(a+b+c\right)}+\dfrac{135}{4\left(a+b+c\right)}\right]\\ \ge\dfrac{3\sqrt{17}}{2}\)

\(S=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)

\(\sqrt{a^2+\dfrac{1}{b^2}}=\dfrac{1}{\sqrt{17}}\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(1+4^2\right)}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\left(1\right)\)\(\left(bunhia\right)\)

\(tương-tự\Rightarrow\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\left(2\right)\)

\(\sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\left(3\right)\)

\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}+b+\dfrac{4}{c}+c+\dfrac{4}{a}\right)\)

\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[16a+\dfrac{4}{a}+16b+\dfrac{4}{b}+16c+\dfrac{4}{c}-15\left(a+b+c\right)\right]\)

\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[2\sqrt{16a.\dfrac{4}{a}}+2\sqrt{16b.\dfrac{4}{b}}+2\sqrt{16c.\dfrac{4}{c}}-15.\dfrac{3}{2}\right]\left(am-gm\right)\)

\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(16+16+16-\dfrac{45}{2}\right)=\dfrac{3\sqrt{17}}{2}\)

\(\Rightarrow MinS=\dfrac{3\sqrt{17}}{2}\Leftrightarrow a=b=c=\dfrac{1}{2}\)

Ta có: \(2\left(b^2+bc+c^2\right)=2b^2+2c^2+2bc\le2b^2+2c^2+b^2+c^2=3\left(b^2+c^2\right)\Rightarrow b^2+c^2\le3-a^2\Rightarrow a^2+b^2+c^2\le3\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\).

Áp dụng bđt Schwars ta có:

\(T\ge a+b+c+\dfrac{18}{a+b+c}=\left(a+b+c+\dfrac{9}{a+b+c}\right)+\dfrac{9}{a+b+c}\ge2\sqrt{9}+\dfrac{9}{3}=9\).

Đẳng thức xảy ra khi a = b = c = 1.

\(không\) \(dùng\) \(bđt\) \(làm\) \(sao\) \(ra\) \(được\) ??

\(\sqrt{a^2+\dfrac{1}{b^2}}=\dfrac{1}{\sqrt{17}}.\sqrt{\left(1+4^2\right)\left(a^2+\dfrac{1}{b^2}\right)}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\left(bunhiacopki\right)\)

\(tương-tự:\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\)

\(\sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\)

\(\Rightarrow Q\ge\dfrac{1}{\sqrt{17}}\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{\sqrt{17}}\left[16a+\dfrac{4}{a}+16b+\dfrac{4}{b}+16c+\dfrac{4}{c}-15\left(a+b+c\right)\right]\)

\(bđt:cosi\Rightarrow16a+\dfrac{4}{a}\ge2\sqrt{16a.\dfrac{4}{a}}=2\sqrt{16.4}=16\)

\(tương-tự\Rightarrow16b+\dfrac{4}{b}\ge16;16c+\dfrac{4}{c}\ge16\)

\(có:a+b+c\le\dfrac{3}{2}\Rightarrow15\left(a+b+c\right)\le\dfrac{45}{2}\)

\(\Rightarrow-15\left(a+b+c\right)\ge-\dfrac{45}{2}\)

\(\Rightarrow Q\ge\dfrac{1}{\sqrt{17}}\left(16+16+16-\dfrac{45}{2}\right)=\dfrac{3\sqrt{17}}{2}\)

\(dấu"="xayra\Leftrightarrow a=b=c=\dfrac{1}{2}\)

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 \(a=b=c=\dfrac{1}{2}\)

Cách này 100% dùng Cô-si

Áp dụng Cô-si:

     \(Q\ge3\sqrt[3]{\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(b^2+\dfrac{1}{c^2}\right)\left(c^2+\dfrac{1}{a^2}\right)}}\)

Ta có:

     \(A=\left(a^2+\dfrac{1}{b^2}\right)\left(b^2+\dfrac{1}{c^2}\right)\left(c^2+\dfrac{1}{a^2}\right)\)

         \(=\left(a^2+b^2+c^2\right)+\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+\left(abc\right)^2+\dfrac{1}{\left(abc\right)^2}\)

Áp dụng Cô-si:

     \(a^2+\dfrac{1}{16a^2}\ge\dfrac{1}{2}\)

     Tương tự với các phần còn lại

\(\Rightarrow A\ge\dfrac{3}{2}+\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+\left(abc\right)^2+\dfrac{1}{\left(abc\right)^2}\)

Ta có:

     \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge3\sqrt[3]{\dfrac{1}{\left(abc\right)^2}}\ge3\sqrt[3]{\dfrac{1}{\left[\dfrac{\left(a+b+c\right)^3}{27}\right]^2}}\ge12\) (Cô-si)

     \(\left(abc\right)^2+\dfrac{1}{64^2\left(abc\right)^2}\ge\dfrac{1}{32}\) (Cô-si)

\(\Rightarrow A\ge\dfrac{3}{2}+\dfrac{15}{16}.12+\dfrac{1}{32}+\dfrac{4095}{64^2\left(abc\right)^2}\)

Mà:

     \(abc\le\dfrac{\left(a+b+c\right)^3}{27}=\dfrac{1}{8}\)

\(\Rightarrow A\ge\dfrac{3}{2}+\dfrac{15}{16}.12+\dfrac{1}{32}+\dfrac{4095}{64^2.\dfrac{1}{8^2}}=\dfrac{4913}{64}\)

\(\Rightarrow Q\ge3\sqrt[3]{\sqrt{A}}\ge\dfrac{3\sqrt{17}}{2}\)

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)

\(\Rightarrow3.P\ge9\Rightarrow P\ge3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(P\ge\dfrac{3abc}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{c^2+a^2}{b^2+\dfrac{c^2+a^2}{2}}\)

\(P\ge\dfrac{3}{2}+2\left(\dfrac{a^2+b^2}{a^2+c^2+b^2+c^2}+\dfrac{b^2+c^2}{a^2+b^2+a^2+c^2}+\dfrac{a^2+c^2}{a^2+b^2+b^2+c^2}\right)\)

Đặt \(\left(a^2+b^2;b^2+c^2;a^2+c^2\right)=\left(x;y;z\right)\)

\(\Rightarrow P\ge\dfrac{3}{2}+2\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=\dfrac{3}{2}+2\left(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{yz+xy}+\dfrac{z^2}{xz+yz}\right)\)

\(P\ge\dfrac{3}{2}+\dfrac{2\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3}{2}+\dfrac{3\left(xy+yz+zx\right)}{xy+yz+zx}=3+\dfrac{3}{2}=\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)