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Lời giải:
Áp dụng BĐT AM-GM:
\(P=\frac{\sqrt{ab}}{(a+c)+(b+c)}+\frac{\sqrt{bc}}{(b+a)+(c+a)}+\frac{\sqrt{ca}}{(c+b)+(a+b)}\)
\(\leq \underbrace{\frac{\sqrt{ab}}{2\sqrt{(a+c)(b+c)}}+\frac{\sqrt{bc}}{2\sqrt{(b+a)(c+a)}}+\frac{\sqrt{ca}}{2\sqrt{(c+b)(a+b)}}}_{M}(*)\)
Xét:
\(M=\frac{1}{2}\frac{\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}(1)\)
Theo BĐT Bunhiacopxky và AM-GM:
\((\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)})^2\leq (ab+bc+ac)(a+b+b+c+c+a)\)
\(=2(ab+bc+ac)(a+b+c)=2[(a+b)(b+c)(c+a)+abc]\)
\(\leq 2[(a+b)(b+c)(c+a)+\frac{(a+b)(b+c)(c+a)}{8}]=\frac{9}{4}(a+b)(b+c)(c+a)\)
\(\Rightarrow \sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}\leq \frac{3}{2}\sqrt{(a+b)(b+c)(c+a)}(2)\)
Từ \((1);(2)\Rightarrow M\leq \frac{1}{2}.\frac{3}{2}=\frac{3}{4}(**)\)
Từ \((*); (**)\Rightarrow P\leq M\leq \frac{3}{4}\)
Vậy \(P_{\max}=\frac{3}{4}\Leftrightarrow a=b=c\)
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)
CMTT: \(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ac}{b+c}+\dfrac{ac}{b+a}\right)\)
\(\Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{ab}{c+a}+\dfrac{ab}{c+b}+\dfrac{bc}{b+a}+\dfrac{bc}{c+a}+\dfrac{ac}{b+c}+\dfrac{ac}{b+c}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left[\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right]=\dfrac{1}{2}\left(a+b+c\right)=1\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
Lợi dụng Cauchy-Schwarz' inequality ta có:
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{ab+ac+bc+c^2}}\)
\(=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự ta cũng có:
\(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ca}{\sqrt{ca+2b}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(P\le\dfrac{1}{2}\left(\dfrac{ab+bc}{a+c}+\dfrac{bc+ca}{a+b}+\dfrac{ab+ca}{b+c}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{b+c}\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\cdot2=1\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)
Ta có P=\(\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}+\dfrac{bc}{\sqrt{bc+\left(a+b+c\right)a}}+\dfrac{ac}{\sqrt{ac+\left(a+b+c\right)b}}\)
=\(\dfrac{ab}{\sqrt{ab+ac+bc+c^2}}+\dfrac{bc}{\sqrt{bc+ac+ab+a^2}}+\dfrac{ac}{\sqrt{ac+ab+bc+b^2}}\)
=\(\dfrac{ab}{\sqrt{a\left(b+c\right)+c\left(b+c\right)}}+\dfrac{bc}{\sqrt{b\left(a+c\right)+a\left(a+c\right)}}+\dfrac{ac}{\sqrt{c\left(a+b\right)+b\left(a+b\right)}}\)
=\(\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}+\dfrac{bc}{\sqrt{\left(b+a\right)\left(c+a\right)}}+\dfrac{ac}{\sqrt{\left(a+b\right)\left(c+b\right)}}\)
áp dụng bđt Cói ta có:
\(\sqrt{\left(a+c\right)\left(b+c\right)}\)\(\le\)\(\dfrac{2+c}{2}=1+\dfrac{c}{2}\)
\(\sqrt{\left(b+á\right)\left(c+a\right)}\)
Theo BĐT \(AM-GM\) ta có :
\(\dfrac{a}{\sqrt{2b^2+2c^2-a^2}}=\dfrac{\sqrt{3}a^2}{\sqrt{3a^2\left(2b^2+2c^2-a^2\right)}}\ge\dfrac{\sqrt{3}a^2}{\dfrac{2a^2+2b^2+2c^2}{2}}=\dfrac{\sqrt{3}a^2}{a^2+b^2+c^2}\)
Tương tự ta có :
\(\dfrac{b}{\sqrt{2c^2+2a^2-b^2}}\ge\dfrac{\sqrt{3}b^2}{a^2+b^2+c^2}\)
\(\dfrac{c}{\sqrt{2a^2+2b^2-c^2}}\ge\dfrac{\sqrt{3}c^2}{a^2+b^2+c^2}\)
Cộng từng vế BĐT :
\(\Rightarrow VT\ge\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=\sqrt{3}\)
\("="\Leftrightarrow a=b=c\)
bai nay t lam roi vao trang chu cua nick thangbnsh cua t keo xuong tim la thay
Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
ad bunhiacopxki ta có
A^2 \(\le3\left(\dfrac{a}{b+c+2a}+\dfrac{b}{c+a+2c}+\dfrac{c}{a+b+2c}\right)\)
Đặt B=\(\dfrac{a}{b+c+2a}+\dfrac{b}{c+a+2b}+\dfrac{c}{a+b+2c}\)
\(\Leftrightarrow\)B-3 =-\(\left(a+b+c\right)\) \(\left(\dfrac{1}{b+c+2a}+\dfrac{1}{c+a+2b}+\dfrac{1}{a+b+2a}\right)\)
dễ CM \(\dfrac{1}{a+b+2c}+\dfrac{1}{b+c+2a}+\dfrac{1}{c+a+2b}\)\(\ge\dfrac{9}{4\left(a+c+b\right)}\)
\(\Rightarrow\)B-3\(\le\)\(\dfrac{-9}{4}\)\(\Rightarrow\)B\(\le\dfrac{3}{4}\)
\(\Rightarrow A^2\le\dfrac{9}{4}\) mà A>0
\(\Rightarrow\)A\(\le\dfrac{3}{2}\)Dấu = xra khi a=b=c
Lời giải:
Đặt biểu thức đã cho là $A$
Ta có:
\(A=\sqrt{\frac{a}{b+c+2a}}+\sqrt{\frac{b}{a+c+2b}}+\sqrt{\frac{c}{a+b+2c}}\)
\(A=\sqrt{\frac{a}{(a+b)+(a+c)}}+\sqrt{\frac{b}{(b+c)+(b+a)}}+\sqrt{\frac{c}{(c+a)+(c+b)}}\)
Áp dụng BĐT AM-GM:
\(A\leq\sqrt{\frac{a}{2\sqrt{(a+b)(a+c)}}}+\sqrt{\frac{b}{2\sqrt{(b+c)(b+a)}}}+\sqrt{\frac{c}{2\sqrt{(c+a)(c+b)}}}\)
\(\Leftrightarrow A\leq \sqrt[4]{\frac{a^2}{4(a+b)(a+c)}}+\sqrt[4]{\frac{b^2}{4(b+c)(b+a)}}+\sqrt[4]{\frac{c^2}{4(c+a)(c+b)}}(*)\)
Tiếp tục áp dụng AM-GM:
\(\sqrt[4]{\frac{a^2}{4(a+b)(a+c)}}\leq \frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{1}{2}+\frac{1}{2}\right)\)
\(\sqrt[4]{\frac{b^2}{4(b+c)(b+a)}}\leq \frac{1}{4}\left(\frac{b}{b+c}+\frac{b}{a+b}+\frac{1}{2}+\frac{1}{2}\right)\)
\(\sqrt[4]{\frac{c^2}{4(c+a)(c+b)}}\leq \frac{1}{4}\left(\frac{c}{c+a}+\frac{c}{c+b}+\frac{1}{2}+\frac{1}{2}\right)\)
Cộng theo vế kết hợp với $(*)$
\(\Rightarrow A\leq \frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+6.\frac{1}{2}\right)\)
\(\Leftrightarrow A\leq \frac{1}{4}.6=\frac{3}{2}\)
Vậy \(A_{\max}=\frac{3}{2}\Leftrightarrow a=b=c\)
\(a=b=c\rightarrow P=\frac{3}{2}\). Ta se c/m do la gtln của P. Thật vậy:
\(\frac{1}{2}P=\sqrt{\frac{1}{4}.\frac{a}{b+c+2a}}+...\)
\(\le\frac{1}{2}\left(\frac{1}{4}+\frac{a}{b+c+2a}+\frac{1}{4}+\frac{b}{c+a+2b}+\frac{1}{4}+\frac{c}{a+b+2c}\right)\)
\(=\frac{1}{2}\left(\frac{3}{4}+\frac{a}{\left(b+a\right)+\left(c+a\right)}+\frac{b}{\left(c+b\right)+\left(b+a\right)}+\frac{c}{\left(c+a\right)+\left(c+b\right)}\right)\)
\(\le\frac{1}{2}\left[\frac{3}{4}+\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)\right]=\frac{3}{4}\)
Do đó \(P\le\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c
Chú ý: \(2a^2+ab+2b^2=\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2\ge\frac{5}{4}\left(a+b\right)^2\) là ok liền:D
Mấy bạn ơi , cho tớ hỏi:
Luật tính điểm hỏi đáp là gì?
Làm thế nào để câu trả lời của mình đứng đầu tiên trong các câu trả lời?
Ai trả lời nhanh mình tích cho.
\(\dfrac{\sqrt{ab}}{a+c+b+c}\le\dfrac{\sqrt{ab}}{2\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{4}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
Tương tự và cộng lại:
\(A\le\dfrac{1}{4}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c\)