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Lời giải tại link sau:
https://hoc24.vn/cau-hoi/cho-abc-la-cac-so-duongcmr-dfrac1a2bcdfrac1b2acdfrac1c2abledfracabc2abc.193908584039
a/ \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\)
\(=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ca+a^2\right)}\)
\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
b/ \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}\)
\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{a+b+c}\)
Dự đoán dấu "=" khi \(a=b=c \Rightarrow P=28\)
Ta sẽ chứng minh \(P=28\) là GTNN
Thật vậy ta có: \(P=\dfrac{ab+bc+ca}{a^2+b^2+c^2}-1+\dfrac{\left(a+b+c\right)^3}{abc}-27\ge0\)
\(\Leftrightarrow\dfrac{ab+bc+ca-\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}+\dfrac{\left(a+b+c\right)^3-27abc}{abc}\ge0\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^3-27abc}{abc}-\dfrac{2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)}{2\left(a^2+b^2+c^2\right)}\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\dfrac{\dfrac{a+b+7c}{2}\cdot\left(a-b\right)^2}{abc}-\dfrac{\left(a-b\right)^2}{2\left(a^2+b^2+c^2\right)}\right)\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{a+b+7c}{2abc}-\dfrac{1}{2\left(a^2+b^2+c^2\right)}\right)\right)\ge0\) *Đúng*
Vậy ...
Lời giải:
Từ \(ab+bc+ac=1\Rightarrow a^2+ab+bc+ac=a^2+1\)
\(\Leftrightarrow (a+b)(a+c)=a^2+1\)
Tương tự: \(\left\{\begin{matrix} b^2+1=(b+c)(b+a)\\ c^2+1=(c+a)(c+b)\end{matrix}\right.\)
Khi đó:
\(A=\frac{(b^2+bc)(c^2+ca)(a^2+ab)}{\sqrt{(a^4+a^2)(b^4+b^2)(c^4+c^2)}}\) \(=\frac{b(b+c)c(c+a)a(a+b)}{\sqrt{a^2b^2c^2(a^2+1)(b^2+1)(c^2+1)}}\)
\(=\frac{abc(a+b)(b+c)(c+a)}{abc\sqrt{(a+b)(a+c)(b+c)(b+a)(c+a)(c+b)}}\) \(=\frac{abc(a+b)(b+c)(c+a)}{abc(a+b)(b+c)(c+a)}=1\)
Vậy \(A=1\)
\(P=2\left(\dfrac{4}{2ab+2ac+2bc}+\dfrac{1}{a^2+b^2+c^2}\right)+\dfrac{1}{2}\left(\dfrac{1}{ab+ac+bc}\right)\)
\(\Rightarrow P\ge\dfrac{2.\left(2+1\right)^2}{2ab+2ac+2bc+a^2+b^2+c^2}+\dfrac{1}{2}.\dfrac{1}{\dfrac{\left(a+b+c\right)^2}{3}}\)
\(\Rightarrow P\ge\dfrac{18}{\left(a+b+c\right)^2}+\dfrac{3}{2}.\dfrac{1}{\left(a+b+c\right)^2}=18+\dfrac{3}{2}=\dfrac{39}{2}\)
\(\Rightarrow P_{min}=\dfrac{39}{2}\) khi \(a=b=c=\dfrac{1}{3}\)
Áp dụng bđt AM-GM:
\(M\ge\dfrac{a^3}{a^2+\dfrac{a^2+b^2}{2}+b^2}+\dfrac{b^3}{b^2+\dfrac{b^2+c^2}{2}+c^2}+\dfrac{c^3}{c^2+\dfrac{a^2+c^2}{2}+a^2}\)
\(=\dfrac{a^3}{\dfrac{3}{2}\left(a^2+b^2\right)}+\dfrac{b^3}{\dfrac{3}{2}\left(b^2+c^2\right)}+\dfrac{c^3}{\dfrac{3}{2}\left(c^2+a^2\right)}\)
\(=\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\)
Xét:
\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\)
\(=a-\dfrac{ab^2}{a^2+b^2}+b-\dfrac{b^2c}{b^2+c^2}+c-\dfrac{c^2a}{c^2+a^2}\)
\(\ge a+b+c-\dfrac{ab^2}{2ab}-\dfrac{b^2c}{2bc}-\dfrac{c^2a}{2ac}=a+b+c-\dfrac{a}{2}-\dfrac{b}{2}-\dfrac{c}{2}=\dfrac{a+b+c}{2}=\dfrac{3}{2}\)
\(\Leftrightarrow M\ge1."="\Leftrightarrow a=b=c=1\)
dòng thứ 5 từ dưới lên cái đầu là bc^2 nhé. Cái sau là ca^2
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(VT=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac+ab+bc+ac+a^2+b^2+c^2}+\dfrac{7}{ab+bc+ac}\)
\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\)
Áp dụng bất đẳng thức AM-GM cho 2 số dương:
\(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1^2}{3}=\dfrac{1}{3}\)
Ta có: \(\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)
\(P\ge\dfrac{3abc}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{c^2+a^2}{b^2+\dfrac{c^2+a^2}{2}}\)
\(P\ge\dfrac{3}{2}+2\left(\dfrac{a^2+b^2}{a^2+c^2+b^2+c^2}+\dfrac{b^2+c^2}{a^2+b^2+a^2+c^2}+\dfrac{a^2+c^2}{a^2+b^2+b^2+c^2}\right)\)
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 \(a=b=c\)