cho a,b,c >0 và \(a^2+b^2+c^2=3\) tìm min của biểu thức
\(P=\frac{a^3}{\sqrt{b^2+3}}+\frac{b^3}{\sqrt{c^2+3}}+\frac{c^3}{\sqrt{a^2+3}}\)
Cho a,b,c>0 và a+b+c=1 Tìm min A = \(\frac{a^2}{\sqrt{a+b}}+\frac{b^2}{\sqrt{b+c}}+\frac{c^2}{\sqrt{c+a}}\) Tìm max B = \(\frac{a^2}{\sqrt[3]{3b+c}}+\frac{b^2}{\sqrt[3]{3c+a}}+\frac{c^2}{\sqrt[3]{3a+b}}\)
Câu 1 : áp dụng BĐT SVAC ta có \(A\ge\frac{(a+b+c)^2}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}}=\frac{1.\sqrt{2a+2b+2c}}{\sqrt{2.}(\sqrt{b+c}+\sqrt{a+b}+\sqrt{a+c})}\)
mặt khác lại có \(\frac{\sqrt{2a+2b+2c}}{\sqrt{2}.(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})}\ge\frac{\sqrt{(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})^2}}{\sqrt{2}.\sqrt{3}.(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c})}=\frac{1}{\sqrt{6}}\)theo bđt svac
\(\Rightarrow A\ge\frac{1}{\sqrt{6}}\)dấu bằng xảy ra tại a=b=c=\(\frac{1}{3}\)
Cho a,b,c dương và a^2+b^2+c^2=3
tìm giá trị nhỏ nhất của biểu thức :\(P=\frac{a^3}{\sqrt{b^2+3}}+\frac{b^3}{\sqrt{c^2+3}}+\frac{c^3}{\sqrt{a^2}+3}\)
Ta có: \(2.2.\sqrt{x^2+3}\le x^2+3+4=x^2+7\Leftrightarrow\sqrt{x^2+3}\le\frac{x^2+7}{4}\) (đẳng thức xảy ra khi x = 1.)
Áp dụng BĐT trên ta có:
\(P\ge4\left(\frac{a^3}{b^2+7}+\frac{b^3}{c^2+7}+\frac{c^3}{a^2+7}\right)=4.\left(\frac{a^4}{ab^2+7a}+\frac{b^4}{bc^2+7b}+\frac{c^4}{ca^2+7c}\right)\ge4.\frac{\left(a^2+b^2+c^2\right)^2}{ab^2+bc^2+ca^2+7\left(a+b+c\right)}\)
( Theo BĐT Schwarz)
Áp dụng BĐT Bunhiacopxki với 3 số ta có:
\(\left(ab^2+bc^2+ca^2\right)^2=\left(b.ab+c.bc+a.ca\right)^2\le\left(a^2+b^2+c^2\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\le\left(a^2+b^2+c^2\right)\frac{\left(a^2+b^2+c^2\right)^2}{3}=\frac{\left(a^2+b^2+c^2\right)^3}{3}=\frac{3^3}{3}=9\Rightarrow ab^2+bc^2+ca^2\le3\)
Ta có: \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=9\Rightarrow a+b+c\le3\)
Do đó:
\(P\ge4.\frac{\left(a^2+b^2+c^2\right)^2}{ab^2+bc^2+ca^2+7\left(a+b+c\right)}\ge\frac{4.3^2}{3+7.3}=\frac{3}{2}\)
Xảy ra đẳng thức khi a = b = c = 1.
Vậy min \(P=\frac{3}{2}\) khi a = b = c = 1.
cho a;b;c>0; a+b+c=6 tìm min
\(P=\frac{a}{\sqrt{b^3+b^2}+4}+\frac{b}{\sqrt{b^3+b^2}+a}+\frac{c}{\sqrt{c^3+c^2+4}}...\)
Cho \(a^2+b^2+c^2=3\left(a,b,c>0\right)\)
Tìm Min
A=\(\frac{a^3}{\sqrt{b^2+3}}+\frac{b^3}{\sqrt{c^2+3}}+\frac{c^3}{\sqrt{a^2+3}}\)
ban oi a=b=c la sai
vi tong cac binh phuong cua chung >3
Cho a,b,c dương và \(a^2+b^2+c^2=3\). Tìm giá trị nhỏ nhất của biểu thức \(P=\frac{a^3}{\sqrt{b^2+3}}+\frac{b^3}{\sqrt{c^2+3}}+\frac{c^3}{\sqrt{a^2+3}}\)
\(\frac{a^3}{\sqrt{b^2+3}}+\frac{a^3}{\sqrt{b^2+3}}+\frac{b^2+3}{8}\ge\frac{3}{2}a^2\)\(\Leftrightarrow\)\(\frac{a^3}{\sqrt{b^2+3}}\ge\frac{3}{4}a^2-\frac{1}{16}b^2-\frac{3}{16}\)
\(P=\Sigma\frac{a^3}{\sqrt{b^2+3}}\ge\frac{3}{4}\left(a^2+b^2+c^2\right)-\frac{1}{16}\left(a^2+b^2+c^2\right)-\frac{9}{16}=\frac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
different way
Ta co:
\(\text{ }P=\Sigma_{cyc}\frac{a^3}{\sqrt{b^2+3}}\ge\Sigma_{cyc}\frac{\left(a^2+b^2+c^2\right)^2}{\Sigma_{cyc}a\sqrt{b^2+3}}\ge\frac{9}{\sqrt{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2+9\right)}}=\frac{3}{2}\)
Dau '=' xay ra khi \(a=b=c=1\)
Cho a,b,c > 0 thỏa mãn: ab + bc + ca =3
Tìm Min \(A=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}+\frac{c}{\sqrt{3+c^2}}\)
Có lẽ là BĐT Cô-si
cứ cho a,b,c>0 thì phải nghĩ ngay đến BĐT cô-si
\(A=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}+\frac{c}{\sqrt{3+c^2}}\)
\(=\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+bc+ca+ab}}+\frac{c}{\sqrt{c^2+ca+ab+bc}}\)
\(=\frac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{\sqrt{b}\cdot\sqrt{b}}{\sqrt{\left(b+c\right)\left(a+b\right)}}+\frac{\sqrt{c}\cdot\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
\(=\frac{\sqrt{a}}{\sqrt{a+b}}\cdot\frac{\sqrt{a}}{\sqrt{c+a}}+\frac{\sqrt{b}}{\sqrt{b+c}}\cdot\frac{\sqrt{b}}{\sqrt{a+b}}+\frac{\sqrt{c}}{\sqrt{c+a}}\cdot\frac{\sqrt{c}}{\sqrt{c+b}}\)
\(\le\frac{\frac{a}{a+b}+\frac{a}{c+a}}{2}+\frac{\frac{b}{b+c}+\frac{b}{a+b}}{2}+\frac{\frac{c}{c+a}+\frac{c}{b+c}}{2}\)
\(=\frac{\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}}{2}=\frac{3}{2}\)
Vậy Max A = 3/2 khi a = b = c = 1. (Max not Min)
1,Cho a,b,c>0 thỏa mãn a+b+c=abc.CMR:
\(\frac{bc}{a\left(1+bc\right)}+\frac{ca}{b\left(1+ca\right)}+\frac{ab}{c\left(1+ab\right)}\ge\frac{3\sqrt{3}}{4}\)
2,Cho a,b,c>0 thỏa mãn \(a^2+b^2+c^2=3\)
Tìm GTLN của P= \(\sqrt{\frac{a^2}{a^2+b+c}}+\sqrt{\frac{b^2}{b^2+c+a}}+\sqrt{\frac{c^2}{c^2+a+b}}\)
3,Cho a,b,c>0 thỏa mãn a+b+c=3.
Tìm GTLN của Q= \(2\sqrt{abc}\left(\frac{1}{\sqrt{3a^2+4b^2+5}}+\frac{1}{\sqrt{3b^2+4c^2+5}}+\frac{1}{\sqrt{3c^2+4a^2+5}}\right)\)
4,Cho a,b,c>0.
Tìm GTLN của P= \(\frac{\sqrt{ab}}{c+3\sqrt{ab}}+\frac{\sqrt{bc}}{a+3\sqrt{bc}}+\frac{\sqrt{ca}}{b+3\sqrt{ca}}\)
ko khó nhưng mà bn đăng từng câu 1 hộ mk mk giải giúp cho
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
Đặt: \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)
=> \(P=\frac{xy}{z^2+3xy}+\frac{yz}{x^2+3yz}+\frac{zx}{y^2+3zx}\)
=> \(3P=\frac{3xy}{z^2+3xy}+\frac{3yz}{x^2+3yz}+\frac{3zx}{y^2+3zx}=1-\frac{z^2}{z^2+3xy}+1-\frac{x^2}{x^2+3yz}+1-\frac{y^2}{y^2+3zx}\)
Ta sẽ CM: \(3P\le\frac{9}{4}\)<=> Cần CM: \(\frac{x^2}{x^2+3yz}+\frac{y^2}{y^2+3zx}+\frac{z^2}{z^2+3xy}\ge\frac{3}{4}\)
Có: \(VT\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)
Ta sẽ CM: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{3}{4}\)
<=> \(4\left(x+y+z\right)^2\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)
<=> \(4\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)+9\left(xy+yz+zx\right)\)
<=> \(x^2+y^2+z^2\ge xy+yz+zx\)
Mà đây lại là 1 BĐT luôn đúng => \(3P\le\frac{9}{4}\)=> \(P\le\frac{3}{4}\)
Vậy P max \(=\frac{3}{4}\)<=> \(a=b=c\)
Cho a,b,c là các số thực dương thay đổi và thoả mãn: \(a+b+c=\frac{3\sqrt{3}}{\sqrt{2}}\) . TÌM GTLN CỦA BIỂU THỨC:
\(M=\frac{1}{a^2+b^2+3}+\frac{1}{b^2+c^2+3}+\frac{1}{c^2+a^2+3}\)
We have:
\(M=1-\frac{1}{3}\Sigma_{cyc}\frac{a^2+b^2}{a^2+b^2+3}\)
Consider:
\(\Sigma_{cyc}\frac{a^2+b^2}{a^2+b^2+3}\ge\frac{3}{2}\)
\(VT\ge\frac{\left(\Sigma_{cyc}\sqrt{a^2+b^2}\right)^2}{2\left(a^2+b^2+c^2\right)+9}\)
Prove:
\(\frac{\left(\Sigma_{cyc}\sqrt{a^2+b^2}\right)^2}{2\left(a^2+b^2+c^2\right)+9}\ge\frac{3}{2}\)
\(\Leftrightarrow4\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge2\left(a^2+b^2+c^2\right)+27\)
Consider:
\(\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge\Sigma_{cyc}a^2+\Sigma_{cyc}ab\)
\(\Rightarrow4\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge4\Sigma_{cyc}a^2+4\Sigma_{cyc}ab\)
Now we need to prove:
\(4\Sigma_{cyc}a^2+4\Sigma_{cyc}ab=2\Sigma_{cyc}a^2+27\)
\(\Leftrightarrow2\left(a+b+c\right)^2=27\) (not fail)
\(\Rightarrow M\le\frac{1}{2}\)
Sign '=' happen when \(a=b=c=\sqrt{\frac{3}{2}}\)
cho a; b; c dương thỏa mãn \(a^2+b^2+c^2=3\)
Tìm Min \(P=\frac{a^3}{\sqrt{b^2+3}}+\frac{b^3}{\sqrt{c^2+3}}+\frac{c^3}{\sqrt{a^2+3}}\)
\(P=\frac{a^4}{a\sqrt{b^2+3}}+\frac{b^4}{b\sqrt{a^2+3}}+\frac{c^4}{c\sqrt{a^2+3}}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a\sqrt{b^2+3}+b\sqrt{a^2+3}+c\sqrt{a^2+3}}\)
Ta co : \(2a\sqrt{b^2+3}+2b\sqrt{c^2+3}+2c\sqrt{a^2+3}\le\frac{4a^2+b^2+3}{2}+\frac{4b^2+a^2+3}{2}+\frac{4a^2+c^2+3}{2}\)
=> \(2\left(a\sqrt{b^2+3}+b\sqrt{c^2+3}+c\sqrt{a^2+3}\right)\le\frac{5\left(a^2+b^2+c^2\right)+9}{2}\) = \(\frac{5.3+9}{2}=12\)
=> \(a\sqrt{b^2+3}+b\sqrt{c^2+3}+c\sqrt{a^2+3}\le6\)
=> \(P\ge\frac{\left(a^2+b^2+c^2\right)^2}{6}=\frac{9}{6}=\frac{3}{2}\)
Dấu '' = '' xảy ra khi và chỉ khi : \(\frac{a^2}{\sqrt{b^2+3}}=\frac{b^2}{\sqrt{c^2+3}}=\frac{c^2}{\sqrt{a^2+3}}\)
\(2a=\sqrt{b^2+3};2b=\sqrt{c^2+3};2c=\sqrt{a^2+3}\)
\(a^2+b^2+c^2=3\)
=> a = b= c = 1