Câu trả lời:
Ta có:
a3 + b3 + c3 - 3abc = 1
\(\Leftrightarrow\) (a + b + c)(a2 + b2 + c2 - ab - bc - ca) = 1
\(\forall\) x ∈ R, ta có:
a2 + b2 + c2 - ab - bc - ca = \(\dfrac{1}{2}\)[(a - b)2 + (b - c)2 + (c - a)2] > 0 [Vì (a + b + c)(a2 + b2 + c2 - ab - bc - ca) = 1 nên phải > 0]
=> a + b + c > 0
Ta có:
(a + b + c)(a2 + b2 + c2 - ab - bc - ca) = 1
\(\Leftrightarrow\) P = \(\dfrac{1}{a+b+c}\) + ab + bc + ca
\(\Leftrightarrow\) 2P = \(\dfrac{2}{a+b+c}\) + 2ab + 2bc + 2ca
\(\Leftrightarrow\) 3P = \(\dfrac{1}{a+b+c}+\dfrac{1}{a+b+c}+\left(a+b+c\right)^2\)
Áp dụng BĐT AM-GM cho 3 số dương, ta có:
\(\dfrac{1}{a+b+c}+\dfrac{1}{a+b+c}+\left(a+b+c\right)^2\ge3\)
\(\Leftrightarrow\) P \(\ge1\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}a+b+c=1&\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=1&\end{matrix}\right.\)(Tự giải nha :)))
Chứng minh AM-GM 3 số:
x + y + z ≥ 3\(\sqrt[3]{xyz}\)
\(\Leftrightarrow\) (\(\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}\))(\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}-\sqrt[3]{xy}-\sqrt[3]{yz}-\sqrt[3]{zx}\)) ≥ 0 (lđ \(\forall\) x, y, z > 0)