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Đề bài trá hình học sinh :)))))))))))))))0
\(\left(a+b+c\right)\left(a'+b'+c'\right)\ge\left(\sqrt{a.a'}+\sqrt{b.b'}+\sqrt{a.a'}\right)^2\\
.\)
=> \(\sqrt{\left(a+b+c\right)\left(a'+b'+c'\right)}\ge\left(\sqrt{a.a'}+\sqrt{b.b'}+\sqrt{c.c'}\right)\\
\)
Dấu chính là điều phải chứng minh :))))))))))))
Bài này áp dụng BĐT Bunhiaacopxki ....................................>< .......................... Chúc học tốt <3
\(x^2-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}+2\right)-1=\frac{1}{4}\left(a-\frac{1}{a}\right)^2\)
\(\Rightarrow\sqrt{x^2-1}=\frac{1}{2}\left(a-\frac{1}{a}\right)\)
Tương tự \(\sqrt{y^2-1}=\frac{1}{2}\left(b-\frac{1}{b}\right)\)
\(A=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab-\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab+\frac{1}{ab}-\frac{a}{b}-\frac{b}{a}}\)
\(=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)
b/ \(B=\frac{\left(\sqrt{a+bx}+\sqrt{a-bx}\right)^2}{a+bx-\left(a-bx\right)}=\frac{a+\sqrt{a^2-b^2x^2}}{bx}\)
\(a^2-b^2x^2=a^2-\frac{4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(m^4+2m^2+1\right)-4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(1-m^2\right)^2}{\left(1+m^2\right)^2}\)
\(\Rightarrow B=\left(a+\frac{a\left(1-m^2\right)}{1+m^2}\right).\left(\frac{1+m^2}{2am}\right)=\frac{a+am^2+a-am^2}{2am}=\frac{1}{m}\)
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Bạn tham khảo:
Câu hỏi của Nguyễn Thu Trà - Toán lớp 9 | Học trực tuyến
a, Đặt \(\sqrt[4]{a}=x;\sqrt[4]{b}=y.\)Bất đẳng thức ban đầu trở thành: \(\frac{2x^2y^2}{x^2+y^2}\le xy.\)
ta có : \(x^2+y^2\ge2xy\Rightarrow\frac{2x^2y^2}{x^2+y^2}\le\frac{2x^2y^2}{2xy}=xy.\)(đpcm )
dấu " = " xẩy ra khi x = y > 0
vậy bất đăng thức ban đầu đúng. dấu " = " xẩy ra khi a = b >0
Kết hợp Mincôpxki và C-S:
\(VT\ge\sqrt{\left(\frac{3}{a+b}+\frac{3}{b+c}+\frac{3}{a+c}\right)^2+\left(a+b+c\right)^2}\)
\(VT\ge\sqrt{\left(\frac{27}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}=\sqrt{\frac{405}{4\left(a+b+c\right)^2}+\frac{81}{\left(a+b+c\right)^2}+\left(a+b+c\right)^2}\)
\(VT\ge\sqrt{\frac{405}{12\left(a^2+b^2+c^2\right)}+2\sqrt{\frac{81\left(a+b+c\right)^2}{\left(a+b+c\right)^2}}}=\sqrt{\frac{405}{12.3}+18}=\frac{3\sqrt{13}}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
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Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(a+b+c\right)\left(a'+b'+c'\right)\ge\left(\sqrt{a\cdot a'}+\sqrt{b\cdot b'}+\sqrt{c\cdot c'}\right)^2\)
\(\Leftrightarrow\sqrt{\left(a+b+c\right)\left(a'+b'+c'\right)}\ge\sqrt{a\cdot a'}+\sqrt{b\cdot b'}+\sqrt{c\cdot c'}\)
Hay \(VP\ge VT\)
Dấu "=" xảy ra khi \(\dfrac{a}{a'}=\dfrac{b}{b'}=\dfrac{c}{c'}\)