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1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
\(A=3a^2c^2+bd+3abc+acd=\left(3a^2c^2+3abc\right)+\left(bd+acd\right)=3ac\left(ac+b\right)+d\left(b+ac\right)\\ =\left(3ac+d\right)\left(ac+b\right)\)
\(B=a^2c-a^2d-b^2d+b^2c=a^2\left(c-d\right)-b^2\left(c-d\right)=\left(a^2-b^2\right)\left(c-d\right)\\=\left(a-b\right)\left(a+b\right)\left(c-d\right)\)
\(C=8x^2+4xy-2ax-ay=\left(8x^2+4xy\right)-\left(2ax+ay\right)=4x\left(2x+y\right)-a\left(2x+y\right)\\ =\left(4x-a\right)\left(2x+y\right)\)
\(E=3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=3\left(a-b\right)^2-12c^2\\ =3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)
Ta có:
a/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a+2c}{3b+2d}\)
b/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-2a}{-2b}=\dfrac{7c}{7d}=\dfrac{-2a+7c}{-2b+7d}\)
PS: Xong
Áp dụng bất đẳng thức Cauchy–Schwarz dạng Engel ta có :
\(VT\ge\frac{\left(2b+3c+2c+3a+2a+3b\right)^2}{a+b+c}\)
\(=\frac{\left(5a+5b+5c\right)^2}{a+b+c}=\frac{\left[5\left(a+b+c\right)\right]^2}{a+b+c}\)
\(=\frac{25\left(a+b+c\right)^2}{a+b+c}=25\left(a+b+c\right)=VP\)
=> đpcm
Đẳng thức xảy ra <=> a = b = c
\(x^3+3x^2+6x+4=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)\)
\(=\left(x+1\right)x^2+2x.\left(x+1\right)+4.\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
a) \(x^3+3x^2+6x+4\)
\(=\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
b) \(3a^2c^2+bd+3abc+acd\)
\(=\left(3a^2c^2+acd\right)+\left(3abc+bd\right)\)
\(=ac\left(3ac+d\right)+b\left(3ac+d\right)\)
\(=\left(ac+b\right)\left(d+3ac\right)\)
Đáng nhẽ phải: ở phân số thứ nhất ở sau chữ chứng minh, trên tử phải là 2x2 + 3ab + 5b2 mới đúng
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
