Câu b,

Ta có BĐT Cauchy \(a^2+b^2\ge2ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Rightarrow\dfrac{ab}{a+b}\le\dfrac{\left(a+b\right)^2}{4\left(a+b\right)}=\dfrac{a+b}{4}\)

Tương tự \(\dfrac{bc}{b+c}\le\dfrac{b+c}{4}\)

\(\dfrac{ac}{a+c}\le\dfrac{a+c}{4}\)

Cộng theo vế ta đc \(VT\le\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)

Dấu bằng xảy ra khi a=b=c

Áp dụng BĐT Cauchy ta có

\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge a\)

\(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\)

\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)

Dấu bằng xảy ra khi a=b=c

Làm tắt vài chỗ thông cảm

Cho x,y là số thực thay đổi thỏa mãn \(\left(x+y\right)^3+4xy\ge2\)

Tìm GTNN \(A=3\left(x^4+y^4+x^2y^2\right)-2\left(x^2+y^2\right)+1\)

1/ Cho a,b,c,d >0 Tìm GTNN

\(A=\dfrac{a}{2b+9c+1945d}+\dfrac{b}{2c+9d+1945a}+\dfrac{c}{2d+9a+1945b}+\dfrac{d}{2a+9b+1945c}\)

2/Cho a;b;c>0 Tm a+b+c=1 Tìm GTNN

\(B=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\)

Giups đc bài nào giúp nha! Thanks