Ta có: \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge0\)

\(\Leftrightarrow1-abc+\left(ab+bc+ca\right)-\left(a+b+c\right)\ge0\)

\(\Leftrightarrow a+b+c-\left(ab+bc+ca\right)\le1-abc\le1\)

Mà \(\hept{\begin{cases}b\ge b^2\\c\ge c^3\end{cases}}\)

\(\Rightarrow a+b^2+c^3-ab-bc-ca\le a+b+c-\left(ab+bc+ca\right)\le1-abc\le1\)

Lời giải:

a.

Vì $\widehat{BAH}=\widehat{CAM}$ nên $\widehat{BAM]=\widehat{CAH}$

Ta có:

\(\frac{HB}{HC}=\frac{S_{BAH}}{S_{CAH}}=\frac{BA.AH.\sin \widehat{BAH}}{CA.AH.\sin \widehat{CAH}}=\frac{AB}{AC}.\frac{\sin \widehat{CAM}}{\sin \widehat{BAM}}(1)\)

\(1=\frac{BM}{CM}=\frac{S_{BAM}}{S_{CAM}}=\frac{AB.AM\sin \widehat{BAM}}{AC.AM.\sin \widehat{CAM}}=\frac{AB.\sin \widehat{BAM}}{AC\sin \widehat{CAM}}\)

\(\Rightarrow \frac{\sin \widehat{CAM}}{\sin \widehat{BAM}}=\frac{AB}{AC}(2)\)

Từ $(1);(2)\Rightarrow \frac{HB}{HC}=\frac{AB^2}{AC^2}$

b.

Đặt $AB=c; BC=a; CA=b$ thì theo phần a ta có:

$\frac{BH}{CH}=\frac{c^2}{b^2}\Rightarrow \frac{BH}{a}=\frac{c^2}{b^2+c^2}$

$\Rightarrow BH=\frac{ac^2}{b^2+c^2}$
$CH=\frac{ab^2}{b^2+c^2}$
Theo định lý Pitago:

$c^2-BH^2=b^2-CH^2$

$\Leftrightarrow c^2-\frac{a^2c^4}{(b^2+c^2)^2}=b^2-\frac{a^2b^4}{(b^2+c^2)^2}$

$\Leftrightarrow (b^2-c^2)=\frac{a^2(b^4-c^4)}{(b^2+c^2)^2}$

$\Leftrightarrow b^2-c^2=\frac{a^2(b^2-c^2)}{b^2+c^2}$

$\Leftrightarrow (b^2-c^2)(b^2+c^2)=a^2(b^2-c^2)$

$\Rightarrow b^2-c^2=0$ hoặc $b^2+c^2=a^2$ 

$\Leftrightarrow AB=AC$ hoặc tam giác $ABC$ vuông tại $A$.

Hình vẽ: