Đăng lâu nhỉ

Ta có:

\(S=pr=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)

\(\Leftrightarrow p^2r^2=p\left(p-a\right)\left(p-b\right)\left(p-c\right)\)

\(\Leftrightarrow r^2=\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p}\)

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

\(\Leftrightarrow\frac{1}{r^2}=4\left(\frac{1}{\left(b+c-a\right)\left(a+c-b\right)}+\frac{1}{\left(a+c-b\right)\left(a+b-c\right)}+\frac{1}{\left(b+c-a\right)\left(a+b-c\right)}\right)\)

\(\Leftrightarrow\frac{1}{4r^2}=\frac{1}{c^2-\left(a-b\right)^2}+\frac{1}{a^2-\left(b-c\right)^2}+\frac{1}{b^2-\left(c-a\right)^2}\)

\(\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)(áp dụng \(x^2-y^2\le x^2\)) 

\(\Rightarrow4r^2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\le1\)

\(\Rightarrow\frac{1}{r^2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}\ge4\left(1\right)\)

Ta lại có

\(S=\frac{a.ha}{2}=pr=\frac{r\left(a+b+c\right)}{2}\)

\(\Rightarrow ha=\frac{r\left(a+b+c\right)}{a}\)

\(\Rightarrow ha^2=\frac{r^2\left(a+b+c\right)^2}{a^2}\)

Tương tự

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

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

Cộng vế theo vế ta được

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

\(\Rightarrow\frac{\left(a+b+c\right)^2}{ha^2+hb^2+hc^2}=\frac{1}{r^2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{\left(a+b+c\right)^2}{ha^2+hb^2+hc^2}\ge4\)

Bài làm này thật xuất sắc !

p và r là gì vậy bạn??

Gọi S là diện tích của tam giác

Ta có : 

\(a=\frac{2S}{h_a};b=\frac{2S}{h_b};c=\frac{2S}{h_c}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\left(a+b+c\right)\left(\frac{h_a+h_b+h_c}{2S}\right)\)

\(=\left(h_a+h_b+h_c\right).\frac{a+b+c}{2S}=\left(h_a+h_b+h_c\right)\left(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right)\)

=> đpcm