Sửa đề: 1+a^2;1+b^2;1+c^2

\(\dfrac{a}{\sqrt{1+a^2}}=\dfrac{a}{\sqrt{a^2+ab+c+ac}}=\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}< =\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

\(\dfrac{b}{\sqrt{1+b^2}}< =\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{b}{b+a}\right)\)

\(\dfrac{c}{\sqrt{1+c^2}}< =\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{a+b}\right)\)

=>\(A< =\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{2}\)

mn giúp em với em đang gấp

Lời giải:

Áp dụng BĐT Cô-si:

$\frac{a^2}{2}+8b^2\geq 2\sqrt{\frac{a^2}{2}.8b^2}=4ab$

$\frac{a^2}{2}+8c^2\geq 2\sqrt{\frac{a^2}{2}.8c^2}=4ac$

$2(b^2+c^2)\geq 2.2\sqrt{b^2c^2}=4bc$

Cộng các BĐT trên theo vế và thu gọn ta được:

$a^2+10(b^2+c^2)\geq 4(ab+bc+ac)=4$

Ta có đpcm.

Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)

Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)

Cộng vế:

\(P\ge\dfrac{a+b+c}{3}=673\)

Dấu "=" xảy ra khi \(a=b=c=673\)

Do \(a^2+b^2+c^2=1\Rightarrow0\le a;b;c\le1\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)\left(b-1\right)\left(c-1\right)\le0\\b^{2011}\le b\\c^{2011}\le c\end{matrix}\right.\)

\(\Rightarrow T\le a+b+c-ab-bc-ca=\left(a-1\right)\left(b-1\right)\left(c-1\right)+1-abc\le1-abc\le1\)

\(T_{max}=1\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị