Xét \(\Delta=\left(2m-1\right)^2-8\left(m-1\right)=4m^2-12m+9=\left(2m-3\right)^2\ge0\)

=> PT luôn có 2 nghiệm x₁,x₂ với mọi m

Theo hệ thức Viet ta có \(\hept{\begin{cases}x_1+x_2=\frac{1-2m}{2}\\x_1x_2=\frac{m-1}{2}\end{cases}}\)

\(\Rightarrow A=\left(x_1+x_2\right)^2-3x_1x_2=\left(\frac{1-2m}{2}\right)^2-\frac{3\left(m-1\right)}{2}\)

\(=\frac{1-4m+4m^2-6m+6}{4}=\frac{4m^2-10m+7}{4}\)

\(=\frac{\left(2m-\frac{5}{2}\right)^2+\frac{3}{4}}{4}\ge\frac{3}{16}\)

Dấu "=" xảy ra khi \(2m=\frac{5}{2}\Rightarrow m=\frac{5}{4}\Rightarrow\frac{a}{b}=\frac{5}{4}\)

\(\Rightarrow4a=5b\Rightarrow2a=\frac{5b}{2}\)

lúc đó \(P=\frac{5b}{2}+2b=\frac{9b}{2}\)

\(LHS\ge\left(\sqrt{ax}.\sqrt{\frac{a}{x}}+\sqrt{bx}.\sqrt{\frac{b}{x}}+\sqrt{cx}.\sqrt{\frac{c}{x}}\right)^2=\left(a+b+c\right)^2\)

TÌm giá trị nhỏ nhất của \(P=\frac{x+7}{\sqrt{x}+3}\)

Xét dạng tổng quát có: \(\frac{1}{\sqrt{n+1}\left(n+1\right)+n\sqrt{n}}=\frac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)-\sqrt{n\left(n+1\right)}}\)

\(< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}-\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào bài toán ta có: 

\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< 1-\frac{1}{\sqrt{2}}\)

\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

.....

\(\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Cộng vế theo vế =>\(VT< 1-\frac{1}{\sqrt{n+1}}\left(ĐPCM\right)\)