Cho a,b,c > 0
Tìm giá trị nhỏ nhất của : \(A=\dfrac{1}{\sqrt{a^2+b^2+c^2+1}}-\dfrac{2}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)
Cho a,b,c là cái số thực dương thỏa mãn a + b + c = 1 . Tìm giá trị nhỏ nhất của biểu thức : Q = \(\dfrac{\left(1-c\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}+\dfrac{\left(1-a\right)^2}{\sqrt{2\left(c+a\right)^2+ca}}\) + \(\dfrac{\left(1-b\right)^2}{\sqrt{2\left(a+b\right)^2+ab}}\)
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
mong mọi người giúp mình câu này
cho a,b,c >0 có \(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}=1\) tìm giá trị lớn nhất của \(\dfrac{a}{\sqrt{bc\left(a^2+1\right)}}+\dfrac{b}{\sqrt{ca\left(b^2+1\right)}}+\dfrac{c}{\sqrt{ab\left(c^2+1\right)}}\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(P=\sqrt{\dfrac{yz}{x^2+1}}+\sqrt{\dfrac{zx}{y^2+1}}+\sqrt{\dfrac{xy}{z^2+1}}\)
\(P=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}+\sqrt{\dfrac{zx}{y^2+xy+yz+zx}}+\sqrt{\dfrac{xy}{z^2+xy+yz+zx}}\)
\(P=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{zx}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right)+\dfrac{1}{2}\left(\dfrac{z}{y+z}+\dfrac{x}{x+y}\right)+\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{3}{2}\)
\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(a=b=c=\sqrt{3}\)
Cho ba số thực dương a,b,c thỏa mãn a+b+c ≤ 2 . Tìm giá trị nhỏ nhất của biểu thức : P = \(\dfrac{b\left(a^2+1\right)^2}{a^2\left(b^2+1\right)}+\dfrac{c\left(b^2+1\right)^2}{b^2\left(c^2+1\right)}+\dfrac{a\left(c^2+1\right)^2}{c^2\left(a^2+1\right)}\)
Giúp mình với mình
\(P\ge3\sqrt[3]{\dfrac{abc\left(a^2+1\right)^2\left(b^2+1\right)^2\left(c^2+1\right)^2}{a^2b^2c^2\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}}=3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)
\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(\dfrac{a+b+c}{3}\right)^3}}=9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(a+b+c\right)^3}}\ge9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{2\left(a+b+c\right)^2}}\)
Theo nguyên lý Dirichlet, trong 3 số \(a^2;b^2;c^2\) luôn có ít nhất 2 số cùng phía so với \(\dfrac{4}{9}\)
Không mất tính tổng quát, giả sử đó là \(a^2;b^2\)
\(\Rightarrow\left(a^2-\dfrac{4}{9}\right)\left(b^2-\dfrac{4}{9}\right)\ge0\)
\(\Leftrightarrow a^2b^2+\dfrac{16}{81}\ge\dfrac{4}{9}a^2+\dfrac{4}{9}b^2\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\dfrac{13}{9}a^2+\dfrac{13}{9}b^2+\dfrac{65}{81}\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\left(c^2+1\right)\)
\(=\dfrac{13}{9}\left(a^2+b^2+\dfrac{4}{9}+\dfrac{1}{9}\right)\left(\dfrac{4}{9}+\dfrac{4}{9}+c^2+\dfrac{1}{9}\right)\)
\(\ge\dfrac{13}{9}\left(\dfrac{2}{3}a+\dfrac{2}{3}b+\dfrac{2}{3}c+\dfrac{1}{9}\right)^2\)
\(\Rightarrow P\ge9\sqrt[3]{\dfrac{\dfrac{13}{9}\left(\dfrac{2}{3}\left(a+b+c\right)+\dfrac{1}{9}\right)^2}{2\left(a+b+c\right)^2}}=9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9\left(a+b+c\right)}\right)^2}\)
\(P\ge9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9.2}\right)^2}=\dfrac{13}{2}\)
\(P_{min}=\dfrac{13}{2}\) khi \(a=b=c=\dfrac{2}{3}\)
Từ giả thiết \(2\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le\dfrac{2}{3}\)
\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)
Đặt \(Q=\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}\)
\(=\dfrac{a^2b^2c^2+\left(a^2b^2+b^2c^2+c^2a^2\right)+\left(a^2+b^2+c^2\right)+1}{abc}\)
\(\ge\dfrac{a^2b^2c^2+3\sqrt[3]{\left(a^2b^2c^2\right)^2}+3\sqrt[3]{a^2b^2c^2}+1}{abc}=\dfrac{\left(\sqrt[3]{a^2b^2c^2}+1\right)^3}{abc}\)
\(=\left(\dfrac{\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{abc}}+\dfrac{1}{\sqrt[3]{abc}}\right)^3=\left(\sqrt[3]{abc}+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)
\(=\left(\sqrt[3]{abc}+\dfrac{4}{9\sqrt[3]{abc}}+\dfrac{5}{9\sqrt[3]{abc}}\right)^3\ge\left(2\sqrt[]{\dfrac{4\sqrt[3]{abc}}{9\sqrt[3]{abc}}}+\dfrac{5}{9.\dfrac{2}{3}}\right)^3=\dfrac{2197}{216}\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{2197}{216}}=\dfrac{13}{2}\)
Cho 3 số thực dương a,b,c thỏa mãn:
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2021\)
Tìm giá trị lớn nhất của P=\(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
ko biết mk làm có đúng ko nhma có gì sai thì đừng trách mk nhé
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{63}{a^2+b^2+c^2}\)
\(6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{a}{ac}\right)+2021\ge\dfrac{54}{ab+bc+ac}+2021\ge\dfrac{54}{a^2+b^2+c^2}+2021\)
<=>\(\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{2021}{9}\)
\(p^2=\left(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\right)^2\)
áp dụng bđt \(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
\(p^2\le3.\left(\dfrac{1}{3\left(2a^2+b^2\right)}+\dfrac{1}{3\left(2b^2+c^2\right)}+\dfrac{1}{3\left(2c^2+a^2\right)}\right)=\dfrac{1}{2a^2+b^2}+\dfrac{1}{2b^2+c^2}+\dfrac{1}{2c^2+a^2}\)
\(< =>p^2\le\dfrac{9}{2a^2+b^2+2b^2+c^2+2c^2+a^2}\)
<=> \(p^2\le3.\dfrac{1}{a^2+b^2+c^2}=\dfrac{2021}{3}< =>p\le\sqrt{\dfrac{2021}{3}}\)
dấu bằng xảy ra khi \(a=b=c=\sqrt{\dfrac{3}{2021}}\)
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2021\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2021\)
\(\Rightarrow2021\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\sqrt{2021.3}=\sqrt{6063}\)
Từ đó:
\(\sqrt{3\left(2a^2+b\right)}=\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}\ge\sqrt{\left(2a+b\right)^2}=2a+b\)
\(\Rightarrow\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}\le\dfrac{1}{2a+b}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\)
Tương tự: \(\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(\Rightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{\sqrt{6063}}{3}\)
\(P_{max}=\dfrac{\sqrt{6063}}{3}\) khi \(a=b=c=\dfrac{3}{\sqrt{6063}}\)
1.cho a,b,c là các số dương thảo man: a+b+c=1. Tìm giá trị lớn nhất của biểu thức:
Q=\(\dfrac{a\left(b+c\right)}{a+1}+\dfrac{b\left(c+a\right)}{b+1}+\dfrac{c\left(a+b\right)}{c+1}\)
2.cho a,b,c dương thỏa man: a2+b2+c2=1
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Cho a,b,c là các số thực dương thỏa mãn a+b+c=3. Tìm giá trị nhỏ nhất của biểu thức:
\(P=\dfrac{1}{a\left(b^2+bc+c^2\right)}+\dfrac{1}{b\left(c^2+ca+a^2\right)}+\dfrac{1}{c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}\)
\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)
\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)
\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)
\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)
\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)
\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Tìm giá trị lớn nhất , giá trị nhỏ nhất của biểu thức :
a)\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\)
b)B=\(\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\)
c)C=\(-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)
Ai lm đc câu nào thì giúp mk với , cảm ơn !!
\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)
Cho 3 số dương a, b, c thay đổi thỏa mãn: \(a^2+b^2+c^2=3\). Tìm giá trị nhỏ nhất của biểu thức: \(P=2.\left(a+b+c\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(P=2\Sigma a+\Sigma\dfrac{1}{a}=\Sigma a+\Sigma a+\Sigma\dfrac{1}{a}\ge3.\sqrt[3]{\left(\Sigma a\right)^2.\Sigma\dfrac{1}{a}}\)
\(Q=\left(\Sigma a\right)^2.\Sigma\dfrac{1}{a}=\left(3+2\Sigma ab\right).\Sigma\dfrac{1}{a}=3\Sigma\dfrac{1}{a}+4\Sigma a+2\Sigma\dfrac{ab}{c}\ge3\Sigma\dfrac{1}{a}+6\Sigma a=3\left(\Sigma\dfrac{1}{a}+2\Sigma a\right)=3P\)\(\Rightarrow\)\(P\ge3\sqrt[3]{3P}\) \(\Leftrightarrow P^3\ge81P\Leftrightarrow P^2\ge81\left(P>0\right)\Leftrightarrow P\ge9\)
" = " \(\Leftrightarrow a=b=c=1\)
Vì $\large a,b,c \in\mathbb{N^*}$ và $\large a^2+b^2+c^2=3\Rightarrow \left\{\begin{matrix} a<\sqrt{3} & \\ b<\sqrt{3} & \\ c<\sqrt{3} & \end{matrix}\right.$
Ta chứng minh bất đẳng thức phụ sau:
Với $0 <x<\sqrt{3}$ thì $2x+\frac{1}{x} \ge x^2.\frac{1}{2}+\frac{5}{2}(*)$
Thật vậy $(*)$ $\large \Leftrightarrow (x-2)(x-1)^2 \le0$
Do $\large x<\sqrt{3}\Leftrightarrow x<2\Leftrightarrow (x-2)(x-1)^2<0$ (Luôn đúng)
Do đó bất đẳng thức được chứng minh
Dấu $"="$ xảy ra khi $x=1$
Trở lại bài toán:
Áp dụng BĐT $(*)$ ta được:
$\large 2a+\frac{1}{a}+2b+\frac{1}{b}+2c+\frac{1}{c}\ge\frac{1}{2}(a^2+b^2+c^2)+\frac{15}{2}=9$
Do $a^2+b^2+c^2=3$
Vậy $GTNN=9$
Dấu $"="$ xảy ra khi: $a=b=c=1$
Cho 3 số dương a, b, c thay đổi thỏa mãn: \(a^2+b^2+c^2=3\). Tìm giá trị nhỏ nhất của biểu thức: \(P=2.\left(a+b+c\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)