cho a, b, c là 3 số thực dương thỏa mãn a+b+c=1
tìm GTLN \(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\)
Cho a,b,c là 3 số thực dương thỏa mãn a+b+c=1.Tìm giá trị lớn nhất của biểu thức P=\(\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ca}{b+ca}}\)
\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
Tương tự: \(\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\) ; \(\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)\)
Cộng vế với vế:
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Cho a,b,c là ba số thực dương thỏa mãn \(a+b+c=2\). Yìm GTLN của biểu thức
\(P=\dfrac{ab}{\sqrt{ab+2c}}+\dfrac{bc}{\sqrt{bc+2a}}+\dfrac{ca}{\sqrt{ac+2b}}\)
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=ab\cdot\sqrt{\dfrac{1}{a+b}\cdot\dfrac{1}{b+c}}\le ab\cdot\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)=\dfrac{1}{2}\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}\right)\)
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 \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+c\left(a+b+c\right)}}=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự:
\(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\) ; \(\dfrac{ca}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)
Cộng vế:
\(P\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}+\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{ab}{b+c}+\dfrac{ca}{b+c}\right)=\dfrac{1}{2}\left(a+b+c\right)=1\)
\(P_{max}=1\) khi \(a=b=c=\dfrac{2}{3}\)
Biết a, b, c là các số dương thỏa mãn \(ab+bc+ca=1\). Tìm GTLN \(P=\dfrac{2a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(P=\frac{2a}{\sqrt{a^2+ab+bc+ac}}+\frac{b}{\sqrt{b^2+ab+bc+ac}}+\frac{c}{\sqrt{c^2+ab+bc+ac}}\\ =\frac{2a}{\sqrt{(a+b)(a+c)}}+\frac{b}{\sqrt{(b+c)(b+a)}}+\frac{c}{\sqrt{(c+a)(c+b)}}\)
\(\leq \frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{4(b+c)}+\frac{b}{b+a}+\frac{c}{4(c+b)}+\frac{c}{c+a}\)
\(=(\frac{a}{a+b}+\frac{b}{b+a})+(\frac{a}{a+c}+\frac{c}{a+c})+\frac{1}{4}(\frac{b}{b+c}+\frac{c}{b+c})=1+1+\frac{1}{4}=\frac{9}{4}\)
Vậy $P_{\max}=\frac{9}{4}$
Cho a,b,c là ba số thực dương thỏa mãn điều kiện ab+bc+ac=3abc. Chứng minh rằng:
\(\sqrt{\dfrac{ab}{a+b+1}}+\sqrt{\dfrac{bc}{b+c+1}}+\sqrt{\dfrac{ca}{c+a+1}}\ge\sqrt{3}\)
Cho a, b, c là số thực dương thỏa mãn: a+b+c=1. Tìm GTLN của biểu thức: \(P=\sqrt{\dfrac{ab}{c+ab}}+\sqrt{\dfrac{bc}{a+bc}}+\sqrt{\dfrac{ac}{b+ac}}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(P=\sum \sqrt{\frac{ab}{c+ab}}=\sum \sqrt{\frac{ab}{c(a+b+c)+ab}}=\sum \sqrt{\frac{ab}{(c+a)(c+b)}}\)
\(\leq \sum \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)=\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)
Vậy $P_{\max}=\frac{3}{2}$ khi $a=b=c=\frac{1}{3}$
cho a,b,c dương thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\). tìm GTLN của \(P=\dfrac{1}{\sqrt{a^2-ab+b^2}}+\dfrac{1}{\sqrt{b^2-bc+c^2}}+\dfrac{1}{\sqrt{c^2-ca+a^2}}\)
\(a^2-ab+b^2=\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}\left(a-b\right)^2\ge\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow P\le\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho a,b,c là ba số thực dương thỏa mãn điều kiện ab+bc+ac=3abc. Chứng minh rằng:
\(\sqrt{\dfrac{ab}{a+b+1}}+\sqrt{\dfrac{bc}{b+c+1}}+\sqrt{\dfrac{ca}{c+a+1}}\ge\sqrt{3}\)
#Toán lớp 9
cho a,b,c dương thỏa mãn điều kiện a+b+c=2018.tìm GTLN
\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)
cho a,b,c dương thỏa mãn điều kiện a+b+c=2018.tìm GTLN
\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)
\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)
\(=\dfrac{a}{a+\sqrt{a.\left(a+b+c\right)+bc}}+\dfrac{b}{b+\sqrt{b.\left(a+b+c\right)+ca}}+\dfrac{c}{c+\sqrt{c.\left(a+b+c\right)+ab}}\)
\(=\dfrac{a}{a+\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{b+\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{c+\sqrt{c^2+ab+bc+ca}}\)
\(=\dfrac{a\left(\sqrt{a^2+ab+bc+ca}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{b^2+ab+bc+ca}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{c^2+ab+bc+ca}-c\right)}{ab+bc+ca}\)
\(=\dfrac{a\left(\sqrt{\left(a+b\right)\left(a+c\right)}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{\left(b+c\right)\left(b+a\right)}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{\left(c+a\right)\left(c+b\right)}-c\right)}{ab+bc+ca}\)
\(\le\dfrac{a\left(\dfrac{2a+b+c}{2}-a\right)}{ab+bc+ca}+\dfrac{b\left(\dfrac{2b+c+a}{2}-b\right)}{ab+bc+ca}+\dfrac{c\left(\dfrac{2c+b+a}{2}-c\right)}{ab+bc+ca}\)
\(=\dfrac{ab+ac}{2\left(ab+bc+ca\right)}+\dfrac{bc+ba}{2\left(ab+bc+ca\right)}+\dfrac{ca+cb}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{2\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=1\)
\(maxP=1\Leftrightarrow a=b=c=\dfrac{2018}{3}\)