\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)

\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)

Đề bài sai, bạn có thể thử kiểm tra với \(a=1.0001\) và \(b=0.9999\)

Nếu mẫu là bình phương, tức \(A=\dfrac{a^4}{\left(b-1\right)^2}+\dfrac{b^4}{\left(a-1\right)^2}\) thì vẫn làm tương tự:

Ta có:

\(\dfrac{a^4}{\left(b-1\right)^2}+16\left(b-1\right)+16\left(b-1\right)+16\ge4\sqrt[4]{\dfrac{a^4.16^3.\left(b-1\right)^2}{\left(b-1\right)^2}}=32a\)

\(\dfrac{b^4}{\left(a-1\right)^2}+16\left(a-1\right)+16\left(a-1\right)+16\ge32b\)

Cộng vế:

\(A+32\left(a+b\right)-32\ge32\left(a+b\right)\)

\(\Rightarrow A\ge32\)

Ta có:

\(\dfrac{a^4}{\left(b-1\right)^3}+16\left(b-1\right)+16\left(b-1\right)+16\left(b-1\right)\ge32a\)

\(\dfrac{b^4}{\left(a-1\right)^3}+16\left(a-1\right)+16\left(a-1\right)+16\left(a-1\right)\ge32b\)

Cộng vế:

\(A+48\left(a+b\right)-96\ge32\left(a+b\right)\)

\(\Leftrightarrow A\ge96-16\left(a+b\right)\ge96-16.4=32\)

\(A_{min}=32\) khi \(a=b=2\)

Đặt \(\left\{{}\begin{matrix}a+c=x>0\\b+c=y>0\end{matrix}\right.\) \(\Rightarrow xy=1\)

\(A=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{\left(x-y\right)^2}+\dfrac{x^2+y^2}{x^2y^2}\)

\(=\dfrac{1}{\left(x-y\right)^2}+x^2+y^2-2xy+2xy\)

\(=\dfrac{1}{\left(x-y\right)^2}+\left(x-y\right)^2+2\ge2\sqrt{\dfrac{\left(x-y\right)^2}{\left(x-y\right)^2}}+2=4\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(P=\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\)

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

\(\geq \frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{a(b+c)+b(a+c)+c(a+b)}=\frac{(ab+bc+ac)^2}{2(ab+bc+ac)}=\frac{ab+bc+ac}{2}\) (thay $1=abc$)

Mà theo BĐT AM-GM:

\(ab+bc+ac\geq 3\sqrt[3]{(abc)^2}=3\). Do đó:

\(P\geq \frac{ab+bc+ac}{2}\geq \frac{3}{2}\)

Vậy \(P_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)

Cách khác:

Áp dụng BĐT AM-GM:

\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{a^3(b+c)}.\frac{a(b+c)}{4}}=\frac{1}{a}=\frac{abc}{a}=bc\)

Tương tự:

\(\frac{1}{b^3(a+c)}+\frac{b(a+c)}{4}\geq ac\)

\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq ab\)

Cộng theo vế các BĐT trên ta có:

\(P+\frac{ab+bc+ac}{2}\geq ab+bc+ac\)

\(\Rightarrow P\geq \frac{ab+bc+ac}{2}\geq \frac{3\sqrt[3]{(abc)^2}}{2}=\frac{3}{2}\)

Vậy \(P_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)

☘ Áp dụng bất đửng thức AM - GM

\(\Rightarrow A=\left(a+b+1\right)\left(a^2+b^2\right)+\dfrac{4}{a+b}\)

\(\ge\left(a+b+1\right)\times2ab+\dfrac{4}{a+b}\)

\(=2\left(a+b+1\right)+\dfrac{4}{a+b}\)

\(=\left(a+b+\dfrac{4}{a+b}\right)+\left(a+b\right)+2\)

\(\ge4+2\sqrt{ab}+2=8\)

⚠ Tự kết luận nha.