28 nhé bạn

hi mk cũng ra thế nhưng k chắc a,b,c=0,4,2 phải k bạn

có cách giải cụ thể k

bài nè cấp 2 chưa làm đc đâu bạn ạ

Có: \(a+b+c=0\)

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

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

\(\Leftrightarrow2\left(ab+bc+ca\right)=-1\) (do \(a^2+b^2+c^2=1\) )

\(\Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab.bc+2bc.ca+2ca.ab=\dfrac{1}{4}\)

\(\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

\(\Leftrightarrow \left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\dfrac{1}{4}\) (do \(a+b+c=0\))

Lại có: \(M=a^4+b^4+c^4\)

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

\(=1-2\left[\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\right]\) (do \(a^2+b^2+c^2=1\))

\(=1-2.\dfrac{1}{4}\)(do \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\dfrac{1}{4}\))

\(=1-\dfrac{1}{2}=\dfrac{1}{2}\)

Vậy \(M=\dfrac{1}{2}\)

\(\left(8x-4x^2-1\right)\left(x^2+2x+1\right)=4\left(x^2+x+1\right)\)

\(\Leftrightarrow8x^3+16x^2+8x-4x^4-8x^3-4x^2-x^2-2x-1=4x^2+4x+4\)

\(\Leftrightarrow11x^2+6x-4x^4-1=4x^2+4x+4\)

\(\Leftrightarrow11x^2+6x-4x^2-1-4x^2-4x-4=0\)

\(\Leftrightarrow7x^2+2x-4x^4-4=0\)

\(\Leftrightarrow\left(-4x^3-4x^2+3x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(-4x^2-8x-5\right)\left(x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vì \(0\le a,b,c\le2\)nên:

\(abc+\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)

\(\Leftrightarrow abc+2bc-abc+2ac-4c+2ab-4b-4a+8\ge0\)

\(\Leftrightarrow2bc+2ac+2ab-4\left(a+b+c\right)+8\ge0\)

\(\Leftrightarrow2\left(ab+bc+ac\right)-12+8\ge0\)

\(\Leftrightarrow2\left(ab+bc+ac\right)\ge4\)

Do đó: \(a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ac\right)\le3^2-4=5\)

(Dấu "="\(\Leftrightarrow\)(a,b,c) là các hoán vị của (0,1,2))

\(ab+1\le b\Rightarrow a+\dfrac{1}{b}\le1\)

Đặt \(\left(a;\dfrac{1}{b}\right)=\left(x;y\right)\Rightarrow x+y\le1\)

Gọi vế trái của BĐT cần chứng minh là P:

\(P=x+\dfrac{1}{x^2}+y+\dfrac{1}{y^2}=\left(\dfrac{1}{x^2}+8x+8x\right)+\left(\dfrac{1}{y^2}+8y+8y\right)-15\left(x+y\right)\)

\(P\ge3\sqrt[3]{\dfrac{64x^2}{x^2}}+3\sqrt[3]{\dfrac{64y^2}{y^2}}-15.1=9\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) hay \(\left(a;b\right)=\left(\dfrac{1}{2};2\right)\)

\(\dfrac{a^5}{b^3+c^2}+\dfrac{b^3+c^2}{4}+\dfrac{a^4}{2}\ge3\sqrt[3]{\dfrac{a^9.\left(b^3+c^2\right)}{8\left(b^3+c^2\right)}}=\dfrac{3a^3}{2}\)

Tương tự và cộng lại:

\(\Rightarrow M-\dfrac{a^4+b^4+c^4}{2}+\dfrac{a^3+b^3+c^3}{4}+\dfrac{a^2+b^2+c^2}{4}\ge\dfrac{3}{2}\left(a^3+b^3+c^3\right)\)

\(\Rightarrow M\ge\dfrac{a^4+b^4+c^4}{2}+\dfrac{5}{4}\left(a^3+b^3+c^3\right)-\dfrac{3}{4}\)

Mặt khác ta có:

\(\dfrac{1}{2}\left(a^4+b^4+c^4\right)\ge\dfrac{1}{6}\left(a^2+b^2+c^2\right)^2=\dfrac{3}{2}\)

\(\left(a^3+a^3+1\right)+\left(b^3+b^3+1\right)+\left(c^3+c^3+1\right)\ge3\left(a^2+b^2+c^2\right)=9\)

\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge9\Rightarrow a^3+b^3+c^3\ge3\)

\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{15}{4}-\dfrac{3}{4}=...\)