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\(a^3-b^3-c^3=3abc\)
\(\Rightarrow a^3-b^3-c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Mà \(a+b+c\ne0\) (độ dài 3 cạnh của 1 tam giác)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left(a-b\right)^2=0;\left(b-c\right)^2=0;\left(c-a\right)^2=0\)
\(\Rightarrow a=b=c\)
Do đó tam giác ABC là tam giác đều
thực hiện trừ 2 vế ta (vế trái cho vế phải) ta được
(a+b+c).(a^2+b^2+c^2 -ab-bc-ca)=0
nên hoặc a+b+c=0 hoặc nhân tử còn lại bằng 0
mà a,b,c là 3 cạnh 1 tam giác nên a+b+c>0
vậy a^2+b^2+c^2 -ab-bc-bc-ca=0
đặt đa thức đó bằng A
A=0 nên 2xA=0
phân tích thành hằng đẳng thức ta có (a-b)2+(b-c)2+(c-a)2=0
nên a=b=c vậy là tam giác đều
Ta có: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
Dấu bằng xảy ra <=> a+b+c=0 hoặc \(a^2+b^2+c^2-ab-ac-bc=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=> a=b=c
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chu vi = 1 => a+b+c=1
viết lại đẳng thức: a/(a+b+c-a)+ b/(a+b+c-b) + c/(a+b+c-c) = 3/2
<=>a/b+c + b/c+a + c/a+b = 3/2
cộng 3 vào 2 vế rút ra được (a+b+c)(1/a+b + 1/b+c + 1/c+a ) = 9/2
<=>1/(a+b)+1/(b+c)+1/(c+a)=9/2(do a+b+c=1)
Sử dụng bđt Schwarz : 1/(a+b)+1/(b+c)+1/(c+a) >/ (1+1+1)2/2(a+b+c) = 9/2
đẳng thức xảy ra <=> a+b=b+c=c+a <=> a=b=c ta có đpcm
\(\Leftrightarrow ab\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+bc\left(\dfrac{1}{a+c}-\dfrac{1}{a+b}\right)+ca\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)=0\)
\(\Leftrightarrow\dfrac{ab\left(a-b\right)}{\left(b+c\right)\left(a+c\right)}+\dfrac{bc\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{ca\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}=0\)
\(\Leftrightarrow\dfrac{ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\) hay tam giác cân