ở trên  a(a-b)+b(b-c)+c(c-a)+0 suy ra a=b=c

thay vào M=a^3x3-3a^3=3a^2 -3a+5=3a^2+-3a+5

GTNN của M là GTNN của 3a^2-3a+5 là bằng 17/4

Câu hỏi của Trần Thị Thùy Linh 2004 - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo nhé!

Ta có: \(1=a^2+b^2+c^2\ge ab+bc+ca\).

\(P=\dfrac{a^3}{b+2c}+\dfrac{b^3}{c+2a}+\dfrac{c^3}{a+2b}=\dfrac{a^4}{ab+2ca}+\dfrac{b^4}{bc+2ab}+\dfrac{c^4}{ca+2bc}\)

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

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

xin bài này , 5 phút sau làm

nếu ai trả lời trc tao , thì thằng đó tự đăng tự tl 

\(\frac{a^3}{2b+C}+\frac{\left(2b+c\right)}{9}+\frac{1}{3}\ge3\sqrt[3]{\frac{a^3}{27}}=a.\)

\(\frac{b^3}{2c+A}+\frac{\left(2c+a\right)}{9}+\frac{1}{3}\ge b\)

\(\frac{c^3}{2a+b}+\frac{\left(2a+b\right)}{9}+\frac{1}{3}\ge c\)

\(VT+\frac{1}{3}\left(a+b+c\right)+\frac{4}{3}\ge3\)

\(VT+\frac{7}{3}\ge3\Leftrightarrow VT\ge1\)

Min của Vt là 1 , dấu =  " khi x=y=z=1

\(\dfrac{4}{3}=a+2\sqrt{\dfrac{a}{4}.b}+\dfrac{1}{2}\sqrt[3]{\dfrac{a}{2}.2b.8c}\)

\(\dfrac{4}{3}\le a+\dfrac{a}{4}+b+\dfrac{1}{6}\left(\dfrac{a}{2}+2b+8c\right)=\dfrac{4}{3}\left(a+b+c\right)\)

\(\Rightarrow a+b+c\ge1\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{16}{21};\dfrac{4}{21};\dfrac{1}{21}\right)\)

Áp dụng BĐT Cô-si dạng Engel,ta có :

\(P=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}=a^2+b^2+c^2\)

\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\Rightarrow\sqrt{3\left(a^2+b^2+c^2\right)}\ge a+b+c\)

\(\Rightarrow6=a+b+c+ab+bc+ac\le\sqrt{3\left(a^2+b^2+c^2\right)}+a^2+b^2+c^2\)

Đặt \(\sqrt{3\left(a^2+b^2+c^2\right)}=t\Rightarrow a^2+b^2+c^2=\frac{t^2}{3}\)

\(\Rightarrow t+\frac{t^2}{3}\ge6\Leftrightarrow3t+t^2-18\ge0\Leftrightarrow\left(t-3\right)\left(t+6\right)\ge0\)

\(\Rightarrow t-3\ge0\Rightarrow t\ge3\)( vì t + 6 > 0 )

\(\Rightarrow P\ge a^2+b^2+c^2=\frac{t^2}{3}\ge3\)

Vậy GTNN của P là 3 khi a = b = c = 1

\(P\ge\dfrac{3abc}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{c^2+a^2}{b^2+\dfrac{c^2+a^2}{2}}\)

\(P\ge\dfrac{3}{2}+2\left(\dfrac{a^2+b^2}{a^2+c^2+b^2+c^2}+\dfrac{b^2+c^2}{a^2+b^2+a^2+c^2}+\dfrac{a^2+c^2}{a^2+b^2+b^2+c^2}\right)\)

Đặt \(\left(a^2+b^2;b^2+c^2;a^2+c^2\right)=\left(x;y;z\right)\)

\(\Rightarrow P\ge\dfrac{3}{2}+2\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=\dfrac{3}{2}+2\left(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{yz+xy}+\dfrac{z^2}{xz+yz}\right)\)

\(P\ge\dfrac{3}{2}+\dfrac{2\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3}{2}+\dfrac{3\left(xy+yz+zx\right)}{xy+yz+zx}=3+\dfrac{3}{2}=\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)