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\(H=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)
\(\ge\sqrt{\left(\dfrac{3}{2}\right)^2+\dfrac{81}{\left(\dfrac{3}{2}\right)^2}}=\dfrac{3\sqrt{17}}{2}\)
Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT tương tự:
\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Bài 1:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{2ab}+\frac{1}{a^2+b^2}\geq \frac{4}{2ab+a^2+b^2}=\frac{4}{a+b)^2}=4(1)\)
Áp dụng BĐT AM-GM:
\(1=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}\Rightarrow \frac{3}{2ab}\geq 6(2)\)
\(a^4+b^4\geq \frac{(a^2+b^2)^2}{2}\geq \frac{(\frac{(a+b)^2}{2})^2}{2}=\frac{1}{8}\) \(\Rightarrow \frac{a^4+b^4}{2}\geq \frac{1}{16}(3)\)
Từ \((1);(2);(3)\Rightarrow P\geq 4+6+\frac{1}{16}=\frac{161}{16}\)
Vậy \(P_{\min}=\frac{161}{16}\). Dấu bằng xảy ra tại $a=b=0,5$
Bài 2:
Áp dụng BĐT Cauchy-Schwarz:
\(2\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)\geq 2. \frac{4}{x^2+y^2+2xy}=\frac{8}{(x+y)^2}=\frac{9}{2}\)
Áp dụng BĐT AM-GM:
\(\frac{80}{81xy}+5xy\geq 2\sqrt{\frac{80}{81}.5}=\frac{40}{9}\)
\(\frac{4}{3}=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{4}{9}\Rightarrow \frac{1}{81ab}\geq \frac{1}{36}\)
Cộng những BĐT vừa cm được ở trên với nhau:
\(\Rightarrow A\geq \frac{9}{2}+\frac{40}{9}+\frac{1}{36}=\frac{323}{36}\)
Vậy \(A_{\min}=\frac{323}{36}\Leftrightarrow a=b=\frac{2}{3}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(a^2+\frac{1}{b^2}\right)(1+1)\geq (a+\frac{1}{b})^2\)
\(\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{a+\frac{1}{b}}{\sqrt{2}}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{d^2}}+\sqrt{d^2+\frac{1}{a^2}}\geq \frac{1}{\sqrt{2}}(a+b+c+d+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\)
Mặt khác theo BĐT Cauchy:
\(a+\frac{1}{a}\geq 2; b+\frac{1}{b}\geq 2; c+\frac{1}{c}\geq 2; d+\frac{1}{d}\geq 2\)
\(\Rightarrow \text{VT}\geq \frac{1}{\sqrt{2}}.8=4\sqrt{2}\)
Vậy giá trị nhỏ nhất của biểu thức là $4\sqrt{2}$. Dấu bằng xảy ra khi $a=b=c=d=1$
Ta có :\(\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}=\dfrac{1}{\sqrt{\left(4a^2+4ab+b^2\right)+\left(a^2-2ab+b^2\right)}}\)
\(=\dfrac{1}{\sqrt{\left(2a+b\right)^2+\left(a-b\right)^2}}\le\dfrac{1}{\sqrt{\left(2a+b\right)^2}}=\dfrac{1}{2a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\) (Cosi)
Tương tự cộng lại ta được :
\(P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{3}\sqrt{3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)}=\dfrac{1}{\sqrt{3}}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\sqrt{3}\)
\(\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)\(\le\) \(\dfrac{1}{3}\sqrt{3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)}\) làm thế nào hả bn ?
Theo giả thiết: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) \(\Leftrightarrow\) \(\dfrac{ab+bc+ca}{abc}=0\)
\(\Leftrightarrow\) ab+bc+ca = 0 \(\Rightarrow\) \(\left\{{}\begin{matrix}ab=-ac-bc\\bc=-ab-ac\\ac=-bc-ab\end{matrix}\right.\)
Xét \(\dfrac{a^2}{a^2+2bc}\) = \(\dfrac{a^2}{a^2+bc-ab-ac}\) = \(\dfrac{a^2}{a\left(a-c\right)-b\left(a-c\right)}\)
= \(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}\)
CMTT \(\dfrac{b^2}{b^2+2ac}\) = \(\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\)
\(\dfrac{c^2}{c^2+2ab}\) = \(\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Rightarrow\) p = \(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}\)+\(\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\)+\(\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
= \(\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\) = 1
a: \(=2ab\cdot\dfrac{-15}{b^2a}=\dfrac{-30}{b}\)
b: \(=\dfrac{2}{3}\cdot\left(1-a\right)=\dfrac{2}{3}-\dfrac{2}{3}a\)
c: \(=\dfrac{\left|3a-1\right|}{\left|b\right|}=\dfrac{3a-1}{b}\)
d: \(=\left(a-2\right)\cdot\dfrac{a}{-\left(a-2\right)}=-a\)
\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)
\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)