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Tách biểu thức như sau:
\(\left(\dfrac{a}{9}+\dfrac{b}{12}+\dfrac{c}{6}+\dfrac{8}{abc}\right)+\left(\dfrac{a}{18}+\dfrac{b}{24}+\dfrac{2}{ab}\right)+\left(\dfrac{b}{16}+\dfrac{c}{8}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{c}{6}+\dfrac{2}{ca}\right)+\left(\dfrac{13a}{18}+\dfrac{13b}{24}\right)+\left(\dfrac{13b}{48}+\dfrac{13c}{24}\right)\)
Đầu tiên em phải dự đoán được điểm rơi (các cặp a;b;c đẹp sao cho \(ab=12\) và \(bc=8\), có các bộ là \(\left(6;2;4\right);\left(3;4;2\right)\)
Sau đó thay 2 bộ kia vào P xem cái nào bằng \(\dfrac{121}{12}\) thì nó đúng (ở đây là 3;4;2)
Khi có điểm rơi, bây giờ chỉ cần tính toán và ghép theo AM-GM để khử tử- mẫu
Cần ghép \(\dfrac{8}{abc}+\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\) (AM-GM 4 số sẽ khử hết biến)
\(\dfrac{8}{abc}=\dfrac{8}{3.4.2}=\dfrac{1}{3}\)
Do đó \(\dfrac{3}{x}=\dfrac{4}{y}=\dfrac{2}{z}=\dfrac{1}{3}\Rightarrow x=9;y=12;z=6\)
Hay ta có bộ đầu tiên: \(\dfrac{a}{9}+\dfrac{b}{12}+\dfrac{c}{6}+\dfrac{8}{abc}\)
Tương tự cho các biến dưới mẫu còn lại, phần dư cuối cùng sẽ ghép cặp a với b (tận dụng \(ab\ge12\)) và b với c, nó sẽ tự đủ
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)
\(\Rightarrow VT=\dfrac{1}{\dfrac{y}{x}\left(\dfrac{z}{y}+1\right)}+\dfrac{1}{\dfrac{z}{y}\left(\dfrac{x}{z}+1\right)}+\dfrac{1}{\dfrac{x}{z}\left(\dfrac{y}{x}+1\right)}\)
\(VT=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)
\(VT\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\dfrac{3}{2}\)
Sửa \(\le\) thành \(\ge\) nha bạn
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
Ta có \(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)
Tương tự: \(\left\{{}\begin{matrix}\dfrac{b^2}{b+ca}=\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}\\\dfrac{c^2}{c+ba}=\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\end{matrix}\right.\)
Áp dụng BĐT cosi:
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3}{4}a\)
\(\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3}{4}b\)
\(\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{c^3}{64}}=\dfrac{3}{4}c\)
Cộng VTV:
\(\Leftrightarrow VT+\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\\ \Leftrightarrow VT\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{2\left(a+b+c\right)}{8}\\ \Leftrightarrow VT\ge\dfrac{a+b+c}{4}\)
Dấu \("="\Leftrightarrow a=b=c=3\)
Lời giải:
$\text{VT}=\frac{a(a+b+c)+bc}{b+c}+\frac{b(a+b+c)+ac}{a+c}+\frac{c(a+b+c)+ab}{a+b}$
$=\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}+\frac{(c+a)(c+b)}{a+b}$
Áp dụng BĐT AM-GM:
$\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}\geq 2\sqrt{(a+b)^2}=2(a+b)$
$\frac{(b+c)(b+a)}{a+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(b+c)^2}=2(b+c)$
$\frac{(a+b)(a+c)}{b+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(c+a)^2}=2(a+c)$
Cộng các BĐT trên theo vế và thu gọn:
$\text{VT}\geq 2(a+b+c)=2$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
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 \(a=b=c=\dfrac{1}{3}\)
Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
Hiển nhiên \(a;b;c\ne0\)
Đặt \(a^2-ab=b^2-bc-c^2-ca=k\ne0\) (do a;b;c phân biệt và khác 0)
\(\Rightarrow\left\{{}\begin{matrix}a-b=\dfrac{k}{a}\\b-c=\dfrac{k}{b}\\c-a=\dfrac{k}{a}\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)+\left(b-c\right)+\left(c-a\right)=\dfrac{k}{a}+\dfrac{k}{b}+\dfrac{k}{c}\)
\(\Rightarrow0=\dfrac{k}{a}+\dfrac{k}{b}+\dfrac{k}{c}\)
\(\Rightarrow k\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=0\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{0}{k}=0\)