A B C M I r D E F

a) Gọi tâm của đường tròn nội tiếp \(\Delta\)ABC là I. (I) tiếp xúc với BC,CA,AB tại D,E,F

Ta có \(S_{BIC}=\frac{1}{2}ID.BC=r.\frac{BC}{2}\). Tương tự \(S_{CIA}=r.\frac{CA}{2};S_{AIB}=r.\frac{AB}{2}\)

Vậy \(S_{ABC}=r.\frac{BC+CA+AB}{2}=pr\)(đpcm).

b) Đặt \(BC=a,CA=b,AB=c,AM=m_A,BM=m_B,CM=m_C\)

Áp dụng công thức tính đường trung tuyến có \(m_A=\frac{\sqrt{2\left(b^2+c^2\right)-a^2}}{2}\)

\(\Rightarrow\frac{1}{m_A}=\frac{2}{\sqrt{2\left(b^2+c^2\right)-a^2}}\), Hoàn toàn tương tự đối với \(m_B,m_C\)

Từ đó \(\frac{1}{m_A}+\frac{1}{m_B}+\frac{1}{m_C}=\frac{2}{\sqrt{2\left(b^2+c^2\right)-a^2}}+\frac{2}{\sqrt{2\left(c^2+a^2\right)-b^2}}+\frac{2}{\sqrt{2\left(a^2+b^2\right)-c^2}}\)

Lại có \(r=\frac{S}{p}=\frac{\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}}{p}=\sqrt{\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p}}\)(Công thức Heron)

\(=\frac{\sqrt{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}{2\sqrt{a+b+c}}\)

Kết hợp với giả thiết \(\frac{1}{m_A}+\frac{1}{m_B}+\frac{1}{m_C}=\frac{1}{r}\) suy ra:

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

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

Áp dụng BĐT Cauchy: \(VT_{\left(1\right)}\le\frac{1}{\sqrt{\left(b+c\right)^2-a^2}}+\frac{1}{\sqrt{\left(c+a\right)^2-b^2}}+\frac{1}{\sqrt{\left(a+b\right)^2-c^2}}\)

\(=\frac{1}{\sqrt{a+b+c}}.\frac{\sqrt{\left(a+b-c\right)\left(b+c-a\right)}+\sqrt{\left(b+c-a\right)\left(c+a-b\right)}+\sqrt{\left(c+a-b\right)\left(a+b-c\right)}}{\sqrt{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}\)\(\le\frac{1}{\sqrt{a+b+c}}.\frac{a+b+c}{\sqrt{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}\)

\(=\frac{\sqrt{a+b+c}}{\sqrt{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}=VP_{\left(1\right)}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c\)<=> \(\Delta\)ABC đều (đpcm).

\(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}=8\)

\(\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2=8\)

\(\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)=8\)

\(18-6\sqrt{5}+6\sqrt{5}-10=8\)

8=8 ( luôn đúng )

\(C=\frac{\sqrt{x+3}}{\sqrt{x-2}}\)\(đkxđ\Leftrightarrow\hept{\begin{cases}x-2\ge0\\x-2\ne0\end{cases}}\)

\(\Rightarrow x-2>0\Rightarrow x>2\)

\(D=\frac{2\sqrt{x}-1}{\sqrt{x}+3}\)\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}+3\ne0\left(tm\right)\end{cases}}\)

\(\Rightarrow x\ge0\)

\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)

\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)

\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)

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

=(x^8-x^7+x^6)-(x^6-x^5+x^4)-(x^3-x^2+x)+(x^2-x+1)

=x^6(x^2-x+1)-x^4(x^2-x+1)-x(x^2-x+1)+(X^2-x+1)

=(x^6-x^4-x+1)(x^2-x+1)>0(vì x^6-x^4-x+1>0 và x^2-x+1>0)