Do a;b;c là 3 cạnh của 1 tam giác nên: \(\left\{{}\begin{matrix}a+b-c>0\\a+c-b>0\\b+c-a>0\end{matrix}\right.\)

\(A=\left(2ab-a^2-b^2+c^2\right)\left(2ab+a^2+b^2-c^2\right)\)

\(=\left[c^2-\left(a-b\right)^2\right]\left[\left(a+b\right)^2-c^2\right]\)

\(=\left(c+a-b\right)\left(c+b-a\right)\left(a+b-c\right)\left(a+b+c\right)>0\) (đpcm)

a^3+b^3+c^3-3abc=0

=>(a+b)^3+c^3-3ab(a+b)-3abc=0

=>(a+b+c)(a^2+b^2+2ab-ac-bc+c^2-3ab)=0

=>(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0

=>a^2+b^2+c^2-ab-bc-ac=0

=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0

=>(a-b)^2+(b-c)^2+(a-c)^2=0

=>a=b=c

\(a< b+c\Rightarrow a^2< ab+ac\)

Tương tự:\(b^2< bc+ab;c^2< ca+cb\)

Cộng lại có đpcm

Lời giải:

BĐT $\Leftrightarrow abc\geq (a+b-c)(b+c-a)(c+a-b)(*)$

Áp dụng BĐT AM-GM:

$(a+b-c)(b+c-a)\leq \left(\frac{a+b-c+b+c-a}{2}\right)^2=b^2$
$(b+c-a)(c+a-b)\leq \left(\frac{b+c-a+c+a-b}{2}\right)^2=c^2$

$(a+b-c)(a+c-b)\leq \left(\frac{a+b-c+a+c-b}{2}\right)^2=a^2$
Nhân theo vế 3 BĐT trên: 

$[(a+b-c)(b+c-a)(c+a-b)]^2\geq (abc)^2$

$\Rightarrow abc\geq (a+b-c)(b+c-a)(c+a-b)$ (BĐT $(*)$ được cm)

Ta có đpcm.

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

\(A=\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\)

\(2A=\frac{2a}{b+c-a}+\frac{2b}{a+c-b}+\frac{2c}{a+b-c}\)

\(2A=\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}=\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)\ge6\)

\(\Leftrightarrow A\ge3."="\Leftrightarrow a=b=c\)

\(a^3+b^3+c^3-3abc=0\)

\(=>\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)

\(=>\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\)

\(=>\left(a+b+c\right).\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

\(=>\left(a+b+c\right).\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(=>\left(a+b+c\right).\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

Vì a,b,c là độ dài 3 cạnh của tam giác nên a,b,c đều lớn hơn 0

\(=>a+b+c\ne0\)

\(=>a^2+b^2+c^2-ab-bc-ac=0\)

\(=>2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\left(1\right)\)

Vì : \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}=>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\) (với mọi a,b,c)

Để (1) thì \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}=>a=b=c}\)

Vậy tam giác đã cho là tam giác đều