Ta có: a + b + c = 0

\(\Rightarrow\) (a + b + c)² = 0

\(\Leftrightarrow\) a² + b² + c² + 2ab + 2bc + 2ac = 0

\(\Leftrightarrow\) 2009 + 2(ab + bc + ac) = 0

\(\Leftrightarrow\) ab + bc + ac = \(\dfrac{-2009}{2}\)

\(\Leftrightarrow\) (ab + bc + ac)² = \(\left(\dfrac{-2009}{2}\right)^2\)

\(\Leftrightarrow\) a²b² + b²c² + a²c² + 2abc(a + b + c) = \(\left(\dfrac{-2009}{2}\right)^2\)

\(\Leftrightarrow\) a²b² + b²c² + c²a² = \(\left(\dfrac{-2009}{2}\right)^2\)    (Vì a + b + c = 0)

Lại có: a² + b² + c² = 2009

\(\Rightarrow\) (a² + b² + c²)² = 2009²

\(\Leftrightarrow\) a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + c²a²) = 2009²

\(\Leftrightarrow\) a⁴ + b⁴ + c⁴ + 2.\(\dfrac{2009^2}{4}\) = 2009²

\(\Leftrightarrow\) a⁴ + b⁴ + c⁴ = 2009² - \(\dfrac{2009^2}{2}\) = 2018040,5

Chúc bn học tốt!

1, C/m : a^3 + b^3 + c^3 ≥ a^2.căn (bc) + b^2.căn (ac) + c^2.căn (ab) 
Ta có : 2( a^3 + b^3 + c^3 ) = ( a^3 + b^3 + c^3 ) + ( a^3 + b^3 + c^3 ) 
≥ 3abc + a^3 + b^3 + c^3 ( BĐT Côsi ) 
= a^3 + abc + b^3 + abc + c^3 + abc ≥ 2.a^2.căn (bc) + 2.b^2.căn (ac) + 2.c^2.căn (ab) ( BĐT Côsi ) 
=> a^3 + b^3 + c^3 ≥ a^2.căn (bc) + b^2.căn (ac) + c^2.căn (ab) 
Dấu " = " xảy ra khi a = b = c. 


2, C/m : (a^2 + b^2 + c^2)(1/(a + b ) + 1/(b + c) +1/(a + c) ) ≥ (3/2)(a + b + c) ( 1 ) 
Áp dụng BĐT Bunhiacốpxki cho phân số ( :D ) ta được : 
(a^2 + b^2 + c^2)(1/(a + b ) + 1/(b + c) +1/(a + c) ) ≥ (a^2 + b^2 + c^2).[(1+1+1)^2/(a+b+b+c+a+c)] = (a^2 + b^2 + c^2) . 9/[2.(a+b+c)] 
(1) <=> (a^2 + b^2 + c^2) . 9/[2.(a+b+c)] ≥ (3/2)(a + b + c) 
<=> 3(a^2 + b^2 + c^2) ≥ (a + b + c)^2 
<=> a^2 + b^2 + c^2 ≥ ab + bc + ca. 
BĐT cuối đúng nên => đpcm ! 
Dấu " = " xảy ra khi a = b = c. 


3, C/m : a^4 + b^4 + c^4 ≥ (a + b + c)abc 
Ta có : 2( a^4 + b^4 + c^4 ) = (a^4 + b^4 +c^4) + (a^4 + b^4 +c^4) 
≥ ( a^2.b^2 + b^2.c^2 + c^2.a^2 ) + (a^4 + b^4 +c^4) = ( a^4 + b^2.c^2 ) + ( b^4 + c^2.a^2 ) + ( c^4 + a^2.b^2 ) 
≥ 2.a^2.bc + 2.b^2.ca + 2.c^2.ab ( BĐT Côsi ) 
= 2.abc(a + b + c) 
Do đó a^4 + b^4 + c^4 ≥ (a + b + c)abc 
Dấu " = " xảy ra khi a = b = c. 

Đặt \(P=a^2+b^2+c^2+ab+bc+ca\)

\(P=\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{2}\left(a^2+b^2+c^2\right)\)

\(P\ge\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{6}\left(a+b+c\right)^2=6\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Đây nhé! Tích giúp c nha

\(P\le a^2+b^2+c^2+3\sqrt{3\left(a^2+b^2+c^2\right)}=12\)

\(P_{max}=12\) khi \(a=b=c=1\)

Lại có: \(\left(a+b+c\right)^2=3+2\left(ab+bc+ca\right)\ge3\Rightarrow a+b+c\ge\sqrt{3}\)

\(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)

\(\Rightarrow\sqrt{3}\le a+b+c\le3\)

\(P=\dfrac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}+3\left(a+b+c\right)\)

\(P=\dfrac{1}{2}\left(a+b+c\right)^2+3\left(a+b+c\right)-\dfrac{3}{2}\)

Đặt \(a+b+c=x\Rightarrow\sqrt{3}\le x\le3\)

\(P=\dfrac{1}{2}x^2+3x-\dfrac{3}{2}=\dfrac{1}{2}\left(x-\sqrt{3}\right)\left(x+6+\sqrt{3}\right)+3\sqrt{3}\ge3\sqrt{3}\)

\(P_{min}=3\sqrt{3}\) khi \(x=\sqrt{3}\) hay \(\left(a;b;c\right)=\left(0;0;\sqrt{3}\right)\) và hoán vị

thế bạn bt hok