\(P=\dfrac{ab\left(a+b\right)+c\left(a^2+b^2\right)}{abc}=\dfrac{a^2+b^2}{ab}+\dfrac{a+b}{c}=\dfrac{a^2+b^2}{ab}+\dfrac{a+b}{\sqrt{a^2+b^2}}\).

Áp dụng bất đẳng thức AM - GM:

\(P\ge\dfrac{a^2+b^2}{ab}+\dfrac{2\sqrt{ab}}{\sqrt{a^2+b^2}}=\left(\dfrac{a^2+b^2}{ab}+\dfrac{2\sqrt{2ab}}{\sqrt{a^2+b^2}}+\dfrac{2\sqrt{2ab}}{\sqrt{a^2+b^2}}\right)-\dfrac{\left(4\sqrt{2}-2\right)\sqrt{ab}}{\sqrt{a^2+b^2}}\ge3\sqrt[3]{\dfrac{a^2+b^2}{ab}.\dfrac{2\sqrt{2ab}}{\sqrt{a^2+b^2}}.\dfrac{2\sqrt{2ab}}{\sqrt{a^2+b^2}}}-\dfrac{\left(4\sqrt{2}-2\right)\sqrt{ab}}{\sqrt{2ab}}=6-\left(4-\sqrt{2}\right)=2+\sqrt{2}\).

Đẳng thức xảy ra khi và chỉ khi tam giác ABC vuông cân tại A.

Đặt \(\hept{\begin{cases}b+c-a=2x\\c+a-b=2y\\a+b-c=2z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=y+z\\b=x+z\\c=x+y\end{cases}}\)

\(\Rightarrow P=\frac{1}{2}.\left(\frac{4\left(y+z\right)}{x}+\frac{9\left(x+z\right)}{y}+\frac{16\left(x+y\right)}{z}\right)\)

\(=\frac{1}{2}.\left(\left(\frac{4y}{x}+\frac{9x}{y}\right)+\left(\frac{4z}{x}+\frac{16x}{z}\right)+\left(\frac{9z}{y}+\frac{16y}{z}\right)\right)\)

\(\ge\frac{1}{2}.\left(2.2.3+2.2.4+2.3.4\right)=26\)

nỏ biết

\(P=\dfrac{a^2\left(b+c\right)+b^2\left(a+c\right)}{abc}=\dfrac{c\left(a^2+b^2\right)+ab\left(a+b\right)}{abc}\)

\(P=\dfrac{a^2+b^2}{ab}+\dfrac{a+b}{c}=\dfrac{a^2+b^2}{ab}+\dfrac{a+b}{\sqrt{a^2+b^2}}\ge\dfrac{a^2+b^2}{ab}+2\sqrt{\dfrac{ab}{a^2+b^2}}\)

Đặt \(\sqrt{\dfrac{a^2+b^2}{ab}}=x\ge\sqrt{2}\)

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

\(P\ge\left(1-\dfrac{1}{2\sqrt{2}}\right).2+3\sqrt[3]{\dfrac{x^2}{2\sqrt{2}x^2}}=2+\sqrt{2}\)

\(P_{min}=2+\sqrt{2}\) khi \(x=\sqrt{2}\Rightarrow a=b\) hay tam giác vuông cân

Bc 5, ac 8

Vì ∆ A’B’C’ đồng dạng với tam giác ABC nên A′B′AB=A′C′AC=B′C′BCA′B′AB=A′C′AC=B′C′BC  (1)

Thay AB = 3(cm), AC = 7 (cm), BC = 5 (cm) , A’B’ = 4,5 (cm) vào (1)

ta có: 4,5/3=A′C′/7=B′C′/5 (cm)

Vậy: A’C’ =7.4,5/3=10,5=7.4,53=10,5 (cm)

B’C’ =5.4,5/3=7,5 (cm).

làm lại dong cuối:\(A\ge\frac{2}{c}+\frac{4}{b}+\frac{6}{a}\)

Mà:\(2c+b=abc\Rightarrow a=\frac{2c+b}{cb}=\frac{2}{b}+\frac{1}{c}\)

\(\Rightarrow2a=\frac{4}{b}+\frac{2}{c}\)

\(\Rightarrow A\ge2a+\frac{6}{a}\)

Ta có:\(A=\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}\right)+2\left(\frac{1}{b+c-a}+\frac{1}{a+b-c}\right)\)

\(+3\left(\frac{1}{a+c-b}+\frac{1}{a+b-c}\right)\)

\(\ge\frac{2}{c}+\frac{4}{b}+\frac{6}{c}\) (Do a,b,c là 3 cạnh của tam giác nên:\(\hept{\begin{cases}a+b-c>0\\a+c-b>0\\c+b-a>0\end{cases}}\)

\(=\frac{6}{a}+2a\ge4\sqrt{3}\left(cosi\right)\left(a>0\right)\)

Dấu = xảy ra khi:

\(a=b=c=\sqrt{3}\)

xin lỗi các bạn đáp án là\(2\sqrt{3}\)

Ban nen cho phan khac chu khong phai phan giai tri