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Với điều kiện như đề bài
Ta có: \(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}=\frac{b^2-a^2+a^2-c^2}{\left(a+b\right)\left(a+c\right)}=\frac{\left(b-a\right)\left(b+a\right)+\left(a-c\right)\left(a+c\right)}{\left(a+b\right)\left(a+c\right)}=\frac{b-a}{a+c}+\frac{a-c}{a+b}\)
Tướng tự:
\(\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}=\frac{c-b}{b+a}+\frac{b-a}{b+c}\)
\(\frac{a^2-b^2}{\left(c+a\right)\left(c+b\right)}=\frac{a-c}{c+b}+\frac{c-b}{c+a}\)
Em nhớ làm tiếp nhé!
a, Áp dụng \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
Áp dụng \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x,y>0\)
Ta có: \(A=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(2+\frac{4}{a+b}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
b, Áp dụng \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)
Áp dụng \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\forall x,y,z>0\)
Ta có: \(B=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2+\left(1+\frac{1}{c}\right)^2\ge\frac{\left(3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\ge\frac{\left(3+\frac{9}{a+b+c}\right)^2}{3}\ge\frac{\left(3+6\right)^2}{3}=27\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)
* Các BĐT phụ bạn tự CM nha! Chúc bạn học tốt
a) \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)
\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=4\)
b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)
\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)
\(\Leftrightarrow B=x^3-20x^2+18x+69\)
c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)
\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)
d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)
\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
Chúc bạn học tốt !