Câu 2 (Bổ Sung) : Chứng minh tam giác đã cho là tam giác đều

Theo bđt tam giác, ta có : \(\begin{cases}a+b>c\\b+c>a\\a+c>b\end{cases}\) \(\Leftrightarrow\begin{cases}bc+ac>c^2\\ab+ac>a^2\\ab+bc>b^2\end{cases}\)

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

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)< a^2+b^2+c^2+2\left(ab+bc+ac\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)< \left(a+b+c\right)^2\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)< 1\)

\(\Leftrightarrow a^2+b^2+c^2< \frac{1}{2}\)

Ta có: \(abc=b+2c\)

\(\Rightarrow a=\dfrac{b+2c}{bc}\)\(\Rightarrow a=\dfrac{1}{c}+\dfrac{2}{b}\)

Áp dụng bất đẳng thức: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

Ta có: \(\dfrac{3}{b+c-a}+\dfrac{4}{c+a-b}+\dfrac{5}{a+b-c}\)

\(=\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}+2\left(\dfrac{1}{b+c-a}+\dfrac{1}{a+b-c}\right)+3\left(\dfrac{1}{c+a-b}+\dfrac{1}{a+b-c}\right)\ge\dfrac{4}{b+c-a+c+a-b}+2.\dfrac{4}{b+c-a+a+b-c}+3.\dfrac{4}{c+a-b+a+b-c}=\dfrac{4}{2c}+2.\dfrac{4}{2b}+3.\dfrac{4}{2a}=\dfrac{2}{c}+\dfrac{4}{b}+\dfrac{6}{a}=2\left(\dfrac{1}{c}+\dfrac{2}{b}+\dfrac{3}{a}\right)=2\left(a+\dfrac{3}{a}\right)\ge2.2\sqrt{\dfrac{a.3}{a}}=4\sqrt{3}\)

(bất đẳng thức Cauchy cho 2 số dương)

\(ĐTXR\Leftrightarrow a=b=c=\sqrt{3}\)

Ta có : \(\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(c+a-2b\right)^2=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)

\(\Leftrightarrow a^2+b^2+4c^2+2ab-4bc-4ac+b^2+c^2+4a^2+2bc-4ca-4ab+c^2+a^2+4b^2+2ac-4bc-4ab=...\)

\(\Leftrightarrow6a^2+6b^2+6c^2-6\left(ab+bc+ca\right)=a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\)

\(\Leftrightarrow6a^2+6b^2+6c^2-6\left(ab+bc+ca\right)-a^2+2ab-b^2-b^2+2bc-c^2-c^2+2ca-a^2=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\)

\(\Leftrightarrow a=b=c\)

<=> Tam giác đó là tam giác đều .

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