a)  Phải là tính theo $a,b,c$ chứ ko thể theo mỗi $a$ được

Kẻ $AH\bot BC$

Ta có $A{{B}^{2}}=A{{H}^{2}}+B{{H}^{2}}$

$\to A{{B}^{2}}=A{{H}^{2}}+{{\left( BC-CH \right)}^{2}}$

$\to A{{B}^{2}}=\left( A{{H}^{2}}+C{{H}^{2}} \right)+B{{C}^{2}}-2BC.CH$

Trong đó có $\cos C=\dfrac{CH}{AC}\Rightarrow CH=AC.\cos C$

$\to A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}-2BC.AC.\cos C$

$\to {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C$

$\to \cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$

$\to {{\cos }^{2}}C=\dfrac{{{\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}^{2}}}{{{\left( 2ab \right)}^{2}}}$

$\to 1-{{\cos }^{2}}C=\dfrac{{{\left( 2ab \right)}^{2}}-{{\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}^{2}}}{{{\left( 2ab \right)}^{2}}}$

$\to {{\sin }^{2}}C=\dfrac{\left( 2ab+{{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)\left( 2ab-{{a}^{2}}-{{b}^{2}}+{{c}^{2}} \right)}{{{\left( 2ab \right)}^{2}}}$

$\to {{\sin }^{2}}C=\dfrac{\left[ {{\left( a+b \right)}^{2}}-{{c}^{2}} \right]\left[ {{c}^{2}}-{{\left( a-b \right)}^{2}} \right]}{{{\left( 2ab \right)}^{2}}}$

$\to {{\sin }^{2}}C=\dfrac{\left( a+b+c \right)\left( a+b-c \right)\left( c+a-b \right)\left( c-a+b \right)}{{{\left( 2ab \right)}^{2}}}$

$\to \sin C=\dfrac{\sqrt{\left( a+b+c \right)\left( a+b-c \right)\left( c+a-b \right)\left( c-a+b \right)}}{2ab}$

$\to \sin C=\dfrac{4\sqrt{\dfrac{a+b+c}{2}\cdot \dfrac{a+b-c}{2}\cdot \dfrac{c+a-b}{2}\cdot \dfrac{c-a+b}{2}}}{2ab}$

Đặt nữa chu vi $p=\dfrac{a+b+c}{2}$

$\to \sin C=\dfrac{2\sqrt{p\left( p-a \right)\left( p-b \right)\left( p-c \right)}}{ab}$

 

 

Ta có ${{S}_{\Delta ABC}}=\dfrac{1}{2}AH\cdot BC$

Trong đó $\sin C=\dfrac{AH}{AC}\Rightarrow AH=AC.\sin C$

$\to {{S}_{\Delta ABC}}=\dfrac{1}{2}AC.BC.\sin C$

$\to {{S}_{\Delta ABC}}=\dfrac{1}{2}\cdot b\cdot a\cdot \dfrac{2\sqrt{p\left( p-a \right)\left( p-b \right)\left( p-c \right)}}{ab}$

$\to {{S}_{\Delta ABC}}=\sqrt{p\left( p-a \right)\left( p-b \right)\left( p-c \right)}$

 

 

b)  Chứng minh ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}\ge 4\sqrt{3}{{S}_{\Delta ABC}}$

Áp dụng BĐT $xyz\le \dfrac{{{\left( x+y+z \right)}^{3}}}{27}$ với $x=p-a\,;\,y=p-b\,;\,z=p-c$

Ta có $\left( p-a \right)\left( p-b \right)\left( p-c \right)\le \dfrac{{{\left( p-a+p-b+p-c \right)}^{3}}}{27}$

$\to \left( p-a \right)\left( p-b \right)\left( p-c \right)\le \dfrac{{{\left( p+2p-a-b-c \right)}^{3}}}{27}$

$\to \left( p-a \right)\left( p-b \right)\left( p-c \right)\le \dfrac{{{p}^{3}}}{27}$

$\to p\left( p-a \right)\left( p-b \right)\left( p-c \right)\le \dfrac{{{p}^{4}}}{27}$

$\to \sqrt{p\left( p-a \right)\left( p-b \right)\left( p-c \right)}\le \dfrac{{{p}^{2}}}{3\sqrt{3}}$

$\to {{S}_{\Delta ABC}}\le \dfrac{\dfrac{{{\left( a+b+c \right)}^{2}}}{4}}{3\sqrt{3}}$

$\to {{S}_{\Delta ABC}}\le \dfrac{{{\left( a+b+c \right)}^{2}}}{12\sqrt{3}}$

 

Áp dụng bất đẳng thức ${{\left( x+y+z \right)}^{2}}\le 3\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$ với $x=a\,;\,y=b\,;\,z=c$

Ta có ${{S}_{\Delta ABC}}\le \dfrac{3\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}{12\sqrt{3}}$

$\to {{S}_{\Delta ABC}}\le \dfrac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{4\sqrt{3}}$

$\to {{a}^{2}}+{{b}^{2}}+{{c}^{2}}\ge 4\sqrt{3}{{S}_{\Delta ABC}}$

Dấu “=” xảy ra khi $a=b=c\Leftrightarrow \Delta ABC$ là tam giác đều