\(ab+bc+ac\le1\)

Ta có \(a^2+b^2+c^2=1\)

\(\Rightarrow ab+bc+ac\le a^2+b^2+c^2\)

Áp dụng bất đẳng thức Cô - si

\(\Rightarrow\left\{\begin{matrix}a^2+b^2\ge2\sqrt{a^2b^2}=2ab\\b^2+c^2\ge2\sqrt{b^2c^2}=2bc\\a^2+c^2\ge2\sqrt{a^2c^2}=2ac\end{matrix}\right.\)

Cộng theo từng vế

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ac\)

\(\Leftrightarrow1\ge ab+bc+ac\) ( đpcm )

phải là \(ab+bc+ca\le1\) nha bởi vì dấu "=" vẫn xảy ra đó.

+ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)

\(\Leftrightarrow2\left(ab+bc+ca\right)\le2\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow ab+bc+ca\le1\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=b=c\\a^2+b^2+c^2=1\end{matrix}\right.\Leftrightarrow a=b=c=\pm\sqrt{\frac{1}{3}}\)

Do $a^2+b^2+c^2=1$ nên $a^2\le 1$ ,$b^2\le 1$ ,$c^2\le 1$
=>$a\ge -1,b\ge -1,c\ge -1$
=>$(1+a)(1+b)(1+c)\ge 0$
=>$1+a+b+c+ab+bc+ca+abc\ge 0$
Cần chứng minh $1+a+b+c+bc+ac+ab\ge 0$
Ta có $1+a+b+c+ab+bc+ca\ge 0$
<=>$a^2+b^2+c^2+ab+bc+ca+a+b+c\ge 0$
<=>$2a^2$

Do $a^2+b^2+c^2=1$ nên $a^2\le 1$ ,$b^2\le 1$ ,$c^2\le 1$
=>$a\ge -1,b\ge -1,c\ge -1$
=>$(1+a)(1+b)(1+c)\ge 0$
=>$1+a+b+c+ab+bc+ca+abc\ge 0$
Cần chứng minh $1+a+b+c+bc+ac+ab\ge 0$

Ta có $1+a+b+c+ab+bc+ca\ge 0$
<=>$a^2+b^2+c^2+ab+bc+ca+a+b+c\ge 0$
<=>$2a^2$

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

\(VT=\Sigma_{cyc}\frac{1}{a^2-ab+b^2}=\Sigma_{cyc}\frac{abc}{a^2-ab+b^2}=\Sigma_{cyc}\frac{abc}{\left(a-b\right)^2+ab}\)

\(\le\Sigma_{cyc}\frac{abc}{ab}=\Sigma_{cyc}c=a+b+c=VP\)

Đẳng thức xảy ra khi \(a=b=c=1\)

P/s: Mình dùng kí hiệu \(\Sigma_{cyc}\) cho gọn, khi làm bạn tự viết rõ ra.