Đặt k=1+3–√2k=1+32k=\dfrac{1+\sqrt{3}}{2}
Ta có:
F=ab+bc+2ca=1ka.(kb)+1kc.(kb)+2ca≤12k(a2+k2b2)+12k(b2+k2b2)+c2+a2=a2(12k+1)+c2(12k+1)+b2kF=ab+bc+2ca=1ka.(kb)+1kc.(kb)+2ca≤12k(a2+k2b2)+12k(b2+k2b2)+c2+a2=a2(12k+1)+c2(12k+1)+b2kF=ab+bc+2ca \\=\dfrac{1}{k}a.(kb)+\dfrac{1}{k}c.(kb)+2ca \\\leq \dfrac{1}{2k}(a^2+k^2b^2)+\dfrac{1}{2k}(b^2+k^2b^2)+c^2+a^2 \\=a^2(\dfrac{1}{2k}+1)+c^2(\dfrac{1}{2k}+1)+b^2k
Do k=1+3–√2k=1+32k=\dfrac{1+\sqrt{3}}{2} nên 12k+1=k12k+1=k\dfrac{1}{2k}+1=k.
Do đó F≤k(a2+b2+c2)=k=1+3–√2F≤k(a2+b2+c2)=k=1+32F \leq k(a^2+b^2+c^2)=k=\dfrac{1+\sqrt{3}}{2}
Dấu '=' khi a=c=kba=c=kba=c=kb và a+b+c=1a+b+c=1a+b+c=1 bạn tự chỉ ra giá trị a,b,ca,b,ca,b,c nhé