Vì a,b,c là độ dài ba cạnh của một tam giác => a,b,c > 0

Sử dụng HĐT a³ + b³ = ( a + b )³ - 3ab( a + b )

a³ + b³ + c³ = 3abc

⇔ a³ + b³ + c³ - 3abc = 0

⇔ ( a³ + b³ ) + c³ - 3abc = 0

⇔ ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

⇔ [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

⇔ ( a + b + c )[ ( a + b )² - ( a + b ).c + c² ] - 3ab( a + b + c ) = 0

⇔ ( a + b + c )( a² + 2ab + b² - ac - bc + c² - 3ab ) = 0

⇔ ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

⇔ \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

Vì a,b,c > 0 => a + b + c > 0 => a + b + c = 0 không thể xảy ra

Xét trường hợp còn lại ta có :

a² + b² + c² - ab - bc - ac = 0

⇔ 2( a² + b² + c² - ab - bc - ac ) = 2.0

⇔ 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

⇔ ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( a² - 2ac + c² ) = 0

⇔ ( a - b )² + ( b - c )² + ( a - c )² = 0

VT luôn ≥ 0 ∀ a,b,c . Dấu "=" xảy ra khi và chỉ khi a = b = c

Kết hợp với điều kiện => a = b = c > 0

=> Tam giác đó là tam giác đều

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(\left(x^2+4x+x+4\right)\left(x^2+2x+3x+6\right)-24\)

\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(x^2+5x+4=a\) ta có

\(a.\left(a+2\right)-24\)

\(a^2+2a-24\)

\(a^2+6a-4a-24\)

\(a\left(a+6\right)-4\left(a+6\right)\)

\(\left(a+6\right)\left(a-4\right)\)

\(\left(x^2+5x+4+6\right)\left(x^2+5x+4-4\right)\)

\(\left(x^2+5x+10\right)\left(x^2+5x\right)\)

   \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

   Đặt \(x^2+5x+5=a\)

   Suy ra  \(\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

               \(=\left(a+1\right)\left(a-1\right)-24\)

                 \(=a^2-1-24=a^2-25=\left(a-5\right)\left(a+5\right)\)

  Do đó 

         \(\left(a+5\right)\left(a-5\right)=x\left(x^2+5x+10\right)\left(x+5\right)\)

               Vậy \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=x\left(x^2+5x+9\right)\left(x+5\right)\)

\(x+\frac{7}{x}=9\Leftrightarrow\frac{x^2+7}{x}=9\Leftrightarrow x^2+7=9x\)

\(\Leftrightarrow x^2-9x+7=0\) 

Ta có : \(\left(-9\right)^2-4.7=81-28=53\)

\(x_1=\frac{9-\sqrt{53}}{2};x_2=\frac{9+\sqrt{53}}{2}\)

?????

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(a+b+c+1\right)^2=\left(a.1+\frac{1}{\sqrt{2}}.\sqrt{2}\left(b+c\right)+\frac{1}{\sqrt{2}}.\sqrt{2}\right)^2\)\(\le\left(a^2+1\right)\text{[}3+2\left(b+c\right)^2\text{]}\)

Khi đó cần CM BĐT : \(\frac{5}{16}\text{[}3+2\left(b+c\right)^2\text{]}\le\left(b^2+1\right)\left(c^2+1\right)\)

Hay: \(16b^2c^2+6\left(b^2+c^2\right)+1\ge20ab\)

BĐT trên đúng theo BĐT AM-GM: \(16b^2c^2+1\ge8bc,6\left(b^2+c^2\right)\ge12bc\)

Dấu '=' xảy ra khi và chỉ khi a=b=c=1/2

TA CÓ: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Do đó: \(\frac{a^3+b^3+c^3}{4abc}=\frac{3}{4}+\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{4abc}\)

\(=\frac{3}{4}+\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\ge\frac{3}{4}+\frac{1}{4}.\frac{9}{ab+bc+ca}\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=\frac{3}{4}+\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{9}{4}=\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{3}{2}\)

\(\Rightarrow P\ge\frac{1}{30}+\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}+\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}-\frac{3}{2}\)

\(=\frac{-22}{15}+\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}+\frac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}\)

\(\ge\frac{-22}{15}+2\sqrt{\left[\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\right]\left[\frac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}\right]}=\frac{-22}{15}+\frac{2}{15}=\frac{-4}{3}\)

Dấu '=' xảy ra <=> a=b=c

Vậy GTNN của P là -4/3 khi a=b=c