\(a^5+b^2+ab+6\ge3a^2b+6\)

\(\Rightarrow P\le\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{\sqrt{a^2b+2}}+\dfrac{1}{\sqrt{b^2c+2}}+\dfrac{1}{\sqrt{c^2a+2}}\right)\le\sqrt{\dfrac{1}{a^2b+2}+\dfrac{1}{b^2c+2}+\dfrac{1}{c^2a+2}}=\sqrt{Q}\)

\(Q=\dfrac{c}{a+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}=\dfrac{1}{2}\left(1-\dfrac{a}{a+2c}+1-\dfrac{b}{b+2a}+1-\dfrac{c}{c+2b}\right)\)

\(Q=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{a^2}{a^2+2ac}+\dfrac{b^2}{b^2+2ab}+\dfrac{c^2}{c^2+2bc}\right)\)

\(Q\le\dfrac{3}{2}-\dfrac{1}{2}\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)

\(\Rightarrow P\le\sqrt{1}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có BĐT phụ: \(a^5+b^5\ge a^2b^2\left(a+b\right)\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\left(a^2+ab+b^2\right)\ge0\)*đúng*

\(\Rightarrow a^5+b^5+ab\ge a^2b^2\left(a+b\right)+ab=ab\left(ab\left(a+b\right)+1\right)\)

\(\Rightarrow\dfrac{ab}{a^5+b^5+ab}\ge\dfrac{ab}{ab\left(ab\left(a+b\right)+1\right)}=\dfrac{1}{ab\left(a+b\right)+1}\)

\(=\dfrac{c}{abc\left(a+b\right)+c}=\dfrac{c}{a+b+c}\left(abc=1\right)\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT\le\dfrac{a+b+c}{a+b+c}=1=VP\)

Khi \(a=b=c=1\)

Tao Không biết làm

Mài cũng có não mà done

đồ vô tâm :<

Dề sai. Cho \(a=c=0,b=\sqrt{2}\) thì được

\(0+\frac{2}{\sqrt{2}+1}+\frac{1}{3}\approx1,162>1\)

ta có a(1-b) \(\ge\)a²(1-b); b(1-c) \(\ge\)b²(1-c); c(1-a) \(\ge\)c²(1-a)

suy ra (a²+b²+c²)-(a²b+b²c+c²a) \(\le\)a(1-b)+b(1-c)+c(1-a)

=> (a²+b²+c²)-(a²b+b²c+c²a) \(\le\)(a+b+c)-(ab+bc+ca)

mà (1-a)(1-b)(1-c) +abc\(\ge\)0 => 1\(\ge\)(a+b+c)-(ab+bc+ca)

vậy a²+b²+c² \(\le\)1+a²b+b²c+c²a

dấu đẳng thức xảy ra <=> trong 3 số có 1 số bằng 0 và 1 số bằng 1

Ta có: \(a.\left(1-b\right)\ge a^2.\left(1-b\right)\)

          \(b.\left(1-c\right)\ge b^2.\left(1-c\right)\)

          \(c.\left(1-a\right)\ge c^2.\left(1-a\right)\)

Suy ra \(\left(a^2+b^2+c^2\right)-\left(a^2b+b^2c+c^2a\right)\le a.\left(1-b\right)+b.\left(1-c\right)+c.\left(1-a\right)\)

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

Mà \(\left(1-a\right).\left(1-b\right).\left(1-c\right)+abc\ge0\) \(\Rightarrow1\ge\left(a+b+c\right)-\left(ab+bc+ca\right)\)

Vậy \(a^2+b^2+c^2\le1+a^2b+b^2c+c^2a\)

Dấu dẳng thức xảy ra \(\Leftrightarrow\)trong ba số đó có một số bằng 0, một số bằng 1 

Trả lời:

Ta có: \(0\le a,b,c\le1\Rightarrow a.\left(1-a\right).\left(1-b\right)\ge0\)

                                       \(\Leftrightarrow a-ab-a^2+ab\ge0\)

                                       \(\Leftrightarrow a^2b\ge ab-a+a^2\)

Tương tự  \(b^2c\ge bc-b+b^2\)

                 \(c^2a\ge ca-c+c^2\)

\(\Rightarrow a^2b+b^2c+c^2a+1\ge1+ab+bc+ca-a-b-c+a^2+b^2+c^2\)

                                                  \(\ge\left(1-a\right).\left(1-b\right).\left(1-c\right)+abc+a^2+b^2+c^2\)

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

Dấu "=" xảy ra \(\Leftrightarrow\left(a,b,c\right)\in\left\{\left(0,1,1\right),\left(1,0,1\right),\left(1,1,0\right),\left(0,0,1\right),\left(0,1,0\right),\left(1,0,0\right)\right\}\)