Trước hết ta c/m bổ đề sau:

Với mọi số thực dương x;y ta luôn có:

\(x^4+y^4\ge xy\left(x^2+y^2\right)\)

Thật vậy, BĐT đã cho tương đương:

\(x^4-x^3y+y^4-xy^3\ge0\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(x^3-y^3\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\) (luôn đúng)

Áp dụng bổ đề trên ta có:

\(T\le\dfrac{a}{bc\left(b^2+c^2\right)+a}+\dfrac{b}{ac\left(a^2+c^2\right)+b}+\dfrac{c}{ab\left(a^2+b^2\right)+c}\)

\(\Rightarrow T\le\dfrac{a^2}{abc\left(b^2+c^2\right)+a^2}+\dfrac{b^2}{abc\left(a^2+c^2\right)+b^2}+\dfrac{c^2}{abc\left(a^2+b^2\right)+c^2}\)

\(\Rightarrow T\le\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{b^2}{a^2+b^2+c^2}+\dfrac{c^2}{a^2+b^2+c^2}=1\)

\(T_{max}=1\) khi \(a=b=c=1\)

\(A=\sqrt{2b\left(a+1\right)}+\sqrt{2c\left(b+1\right)}+\sqrt{2a\left(c+1\right)}\)

\(A=\dfrac{1}{2\sqrt{2}}.2\sqrt{4b\left(a+1\right)}+\dfrac{1}{2\sqrt{2}}.2\sqrt{4c\left(b+1\right)}+\dfrac{1}{2\sqrt{2}}.2\sqrt{4a\left(c+1\right)}\)

\(A\le\dfrac{1}{2\sqrt{2}}\left(4b+a+1\right)+\dfrac{1}{2\sqrt{2}}\left(4c+b+1\right)+\dfrac{1}{2\sqrt{2}}\left(4a+c+1\right)\)

\(A\le\dfrac{1}{2\sqrt{2}}\left[5\left(a+b+c\right)+3\right]=2\sqrt{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

bài khó thế

Đánh sai đề kìa :v \(\frac{1}{\sqrt{b^2-ab+2a^2}}\) mới đúng. 

Cho \(a=b\rightarrow S=2\sqrt{2}\). Ta cm đây là gtln của S.

\(S\le\left(a+b\right)\sqrt{2\left(\frac{1}{a^2-ab+2b^2}+\frac{1}{b^2-ab+2a^2}\right)}\le2\sqrt{2}\)

\(\Leftrightarrow\left(5a^2-6ab+5b^2\right)\left(a-b\right)^2\ge0\)(bình phương lên quy đồng là xong)

Đẳng thức xảy ra khi a  = b.

Đúng ko đấy ạ, sao em quy đồng lên ra \(20a^2b^2-16\left(a^3b+ab^3\right)+5\left(a^4+b^4\right)\)

Nhưng \(\left(a-b\right)^2\left(5a^2-6ab+5b^2\right)=5\left(a^4+b^4\right)+22a^2b^2-16\left(a^3b+ab^3\right)\)

Anh đã kiểm tra lại bằng Maple. Đúng em nhé, em thử quy đồng lại hoặc vào thống kê hỏi đáp xem ảnh.

Gọi M, N, K, I lần lượt biểu diễn số phức 

Theo đề ra, ta có:

\(a^2+b^2+c^2\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)

\(=a^3+b^3+c^3+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)

Theo BĐT Cô-si:

\(\left\{{}\begin{matrix}a^3+ab^2\ge2a^2b\\b^3+bc^2\ge2b^2c\\c^3+ca^2\ge2c^2a\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge3\left(a^2b+b^2c+c^2a\right)\)

Do vậy \(M\ge14\left(a^2+b^2+c^2\right)+\dfrac{3\left(ab+bc+ac\right)}{a^2+b^2+c^2}\)

Ta đặt \(a^2+b^2+c^2=k\)

Luôn có \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=1\)

Vì thế nên \(k\ge\dfrac{1}{3}\)

Khi đấy:

\(M\ge14k+\dfrac{3\left(1-k\right)}{2k}=\dfrac{k}{2}+\dfrac{27k}{2}+\dfrac{3}{2k}-\dfrac{3}{2}\ge\dfrac{1}{3}.\dfrac{1}{2}+2\sqrt{\dfrac{27k}{2}.\dfrac{3}{2k}}-\dfrac{3}{2}=\dfrac{23}{3}\)

\(\Rightarrow Min_M=\dfrac{23}{3}\Leftrightarrow a=b=c=\dfrac{1}{3}\).

\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)

Tương tự và cộng lại:

\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)

\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)

\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)

Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z

\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)

Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))

\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))

\(\Rightarrow P\le\dfrac{3}{16}\)

\(ĐTXR\Leftrightarrow a=b=c=1\)