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Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(B=\frac{5bc}{a^2b+a^2c}+\frac{5ac}{b^2a+b^2c}+\frac{5ab}{c^2b+c^2a}\)
\(B=5\left(\frac{\frac{1}{a^2}}{\frac{1}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{1}{c}+\frac{1}{a}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\right)\)\(\geq 5\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{c}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{a}+\frac{1}{b}}\)
hay \(B\geq \frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Áp dụng BĐT AM-GM:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\sqrt[3]{\frac{1}{abc}}=3\) do \(abc=1\)
Suy ra \(B\geq \frac{15}{2}\Leftrightarrow B_{\min}=\frac{15}{2}\)
Dấu bằng xảy ra khi \(a=b=c=1\)
\(P=\dfrac{bc}{\dfrac{a^2bc}{c}+\dfrac{a^2bc}{b}}+\dfrac{ca}{\dfrac{b^2ac}{a}+\dfrac{b^2ac}{c}}+\dfrac{ab}{\dfrac{c^2ab}{b}+\dfrac{c^2ab}{a}}=\dfrac{\left(bc\right)^2}{a^2b^2c+a^2bc^2}+\dfrac{\left(ca\right)^2}{b^2a^2c+b^2ac^2}+\dfrac{\left(ab\right)^2}{c^2a^2b+c^2ab^2}=\dfrac{\left(bc\right)^2}{ab+ac}+\dfrac{\left(ca\right)^2}{ba+bc}+\dfrac{\left(ab\right)^2}{ca+cb}\ge\dfrac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}\ge\dfrac{3\sqrt[3]{\left(abc\right)^2}}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 1