Ta chứng minh: \(2\left(a^2-ab+b^2\right)^2\ge b^4+a^4\left(1\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)^2\ge0\)( Luôn đúng \(\forall a;b\))
Tương tự có: \(2\left(b^2-bc+c^2\right)^2\ge b^4+c^4\left(2\right)\)
Và: \(2\left(c^2-ca+a^2\right)^2\ge a^4+c^4\left(3\right)\)
Ta nhân các vế trên ta được: \(8\left(a^2-ab+b^2\right)^2\left(b^2-bc+c^2\right)^2\left(c^2-ca+a^2\right)^2\ge\left(a^4+b^4\right)\left(b^4+c^4\right)\left(c^4+a^4\right)=8\)
Hay: \(\left(a^2-ab+b^2\right)\left(b^2-bc+c^2\right)\left(c^2-ca+a^2\right)\ge1\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Trâu bò:
Giả sử c = min{a,b,c}
Đặt a =x +c; b = y +c;c=c thì x,y >= 0
C/m: \(8\left[\left(a^2-ab+b^2\right)\left(b^2-bc+c^2\right)\left(c^2-ca+a^2\right)\right]^2\ge\left(a^4+b^4\right)\left(b^4+c^4\right)\left(c^4+a^4\right)\)
Xét hiệu hai vế thu được:
\(c*(12*x^3*y^8-8*x^4*y^7+16*x^5*y^6+16*x^6*y^5-8*x^7*y^4+12*x^8*y^3)+c^2*(18*x^2*y^8-16*x^3*y^7+60*x^4*y^6+60*x^6*y^4-16*x^7*y^3+18*x^8*y^2)+c^3*(12*x*y^8+16*x^2*y^7+88*x^4*y^5+88*x^5*y^4+16*x^7*y^2+12*x^8*y)+c^4*(6*y^8+16*x*y^7+32*x^2*y^6-32*x^3*y^5+242*x^4*y^4-32*x^5*y^3+32*x^6*y^2+16*x^7*y+6*x^8)+7*x^4*y^8+c^5*(16*y^7+16*x*y^6+88*x^3*y^4+88*x^4*y^3+16*x^6*y+16*x^7)-16*x^5*y^7+c^6*(24*y^6-16*x*y^5+60*x^2*y^4+60*x^4*y^2-16*x^5*y+24*x^6)+24*x^6*y^6+c^7*(16*y^5-8*x*y^4+16*x^2*y^3+16*x^3*y^2-8*x^4*y+16*x^5)-16*x^7*y^5+c^8*(8*y^4-16*x*y^3+24*x^2*y^2-16*x^3*y+8*x^4)+7*x^8*y^4\)Dấu " * " là nhân.
Dễ thấy nó đúng -> qed
