Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3.
\(\dfrac{2a^2}{b^2}+2\dfrac{b^2}{c^2}+2\dfrac{c^2}{a^2}\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
áp dụng bất đẳng thức cosi
+ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\dfrac{a}{c}\)
......
tương tự với 2 cái sau
Áp dụng bđt Cauchy-Schwarz:
\(P=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\ge\dfrac{\left(1+1\right)^2}{a^2+b^2+2ab}=\dfrac{4}{\left(a+b\right)^2}\ge\dfrac{4}{1^2}=4\)\("="\Leftrightarrow a=b=\dfrac{1}{2}\)
Đặt \(\left(x^3;y^3;z^3\right)=\left(a;b;c\right)\left(x,y,z>0\right)\)
\(\Rightarrow xyz=1\)
Ta cần chứng minh
\(\dfrac{1}{x^3+y^3+1}+\dfrac{1}{y^3+z^3+1}+\dfrac{1}{z^3+x^3+1}\le1\)
Áp dụng AM-GM, ta có: \(x^3+y^3+1=\left(x+y\right)\left(x^2-xy+y^2\right)+xyz\)
\(\ge\left(x+y\right)xy+xyz=xy\left(x+y+z\right)\)
\(\Rightarrow\dfrac{1}{x^3+y^3+1}\le\dfrac{1}{xy\left(x+y+z\right)}\)
Tương tự: \(\dfrac{1}{y^3+z^3+1}\le\dfrac{1}{yz\left(x+y+z\right)}\)
\(\dfrac{1}{z^3+x^3+1}\le\dfrac{1}{zx\left(x+y+z\right)}\)
Cộng vế theo vế, ta được
\(....\le\dfrac{1}{x+y+z}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{1}{x+y+z}.\dfrac{x+y+z}{xyz}=\dfrac{1}{xyz}=1\)
Vậy ta có đpcm
Đẳng thức xảy ra khi a=b=c=1
Từ \(\dfrac{a}{1+a}+\dfrac{2b}{2+b}+\dfrac{3c}{3+c}\le\dfrac{6}{7}\)
\(\Leftrightarrow1-\dfrac{a}{1+a}+2-\dfrac{2b}{2+b}+3-\dfrac{3c}{3+c}\ge6-\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{4}{b+2}+\dfrac{9}{c+3}\ge\dfrac{36}{7}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{1}{a+1}+\dfrac{4}{b+2}+\dfrac{9}{c+3}\)
\(\ge\dfrac{\left(1+2+3\right)^2}{a+b+c+6}=\dfrac{36}{7}=VP\)
Xảy ra khi \(a=\dfrac{1}{6};b=\dfrac{1}{3};c=\dfrac{1}{2}\)
2) \(\dfrac{1}{x}+\dfrac{25}{y}+\dfrac{64}{z}=\dfrac{4}{4x}+\dfrac{225}{9y}+\dfrac{1024}{16z}\ge\dfrac{\left(2+15+32\right)^2}{4x+9y+6z}=49\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Nhận xét: Với x,y > 0 ta có:
\(4xy\le\left(x+y\right)^2\)
<=> \(\dfrac{1}{x+y}\le\dfrac{x+y}{4xy}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Xảy ra khi x = y
Áp dụng và bài ta có:
\(\dfrac{1}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{2a}+\dfrac{1}{b+c}\right)\le\dfrac{1}{4}\left[\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\right]=\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{2c}\right)\)
Tương tự: \(\dfrac{1}{a+2b+c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2c}\right)\);
\(\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{c}\right)\)
Cộng 3 vế bđt có:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{3}{4}\)
\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{c^2+\dfrac{1}{c^2}}\)
\(\Leftrightarrow\sqrt{\dfrac{97}{4}}P=\sqrt{4+\dfrac{81}{4}}\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{c^2+\dfrac{1}{c^2}}\)
\(\ge\left(2a+\dfrac{9}{2a}\right)+\left(2b+\dfrac{9}{2b}\right)+\left(2c+\dfrac{9}{2c}\right)\)
\(=2\left(a+b+c\right)+\dfrac{9}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\ge4+\dfrac{9}{2}.\dfrac{9}{a+b+c}=4+\dfrac{81}{4}=\dfrac{97}{4}\)
\(\Rightarrow P\ge\sqrt{\dfrac{97}{4}}\)
PS: Lần sau chép đề cẩn thận nhé bạn.
Áp dụng BĐT AM-GM ta có:
\(A=a+b+\dfrac{1}{a}+\dfrac{1}{b}\)
\(=\left(a+\dfrac{1}{4a}\right)+\left(b+\dfrac{1}{4b}\right)+3\left(\dfrac{1}{4a}+\dfrac{1}{4b}\right)\)
\(=2\sqrt{a\cdot\dfrac{1}{4a}}+2\sqrt{b\cdot\dfrac{1}{4b}}+3\dfrac{\left(1+1\right)^2}{4\left(a+b\right)}\)
\(\ge2\cdot\dfrac{1}{2}+2\cdot\dfrac{1}{2}+\dfrac{3\cdot4}{4}=5=VP\)
Xảy ra khi \(a=b=\dfrac{1}{2}\)