CHo 3 số thực a,b,c thỏa mãn \(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=3\). chứng minh rằng\(\dfrac{27a^2}{c\left(c^2+9a^2\right)}+\dfrac{b^2}{a\left(4a^2+b^2\right)}+\dfrac{8c^2b}{b\left(9b^2+4c^2\right)}\ge\dfrac{3}{2}\)
Cho 3 số thực dương x,y,z thỏa mãn \(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=3\)
Chứng minh \(\dfrac{27a^2}{c\left(c^2+9a^2\right)}+\dfrac{b^2}{a\left(4a^2+b^2\right)}+\dfrac{8c^3}{b\left(9b^2+4c^2\right)}\ge\dfrac{3}{2}\)
Cho 3 số thực a,b,c thỏa mãn \(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=3\).Chứng minh rằng
\(\dfrac{27a^2}{c\left(c^2+9a^2\right)}+\dfrac{b^2}{a\left(4a^2+b^2\right)}+\dfrac{8c^2}{b\left(9b^2+4c^2\right)}\ge\dfrac{3}{2}\)
Với a, b, c là những số thực dương thỏa mãn \(\left(a+b\right)\left(b+c\right)\)\(\left(c+a\right)\)=1
Chứng minh rằng \(\dfrac{a}{b\left(b+2c\right)^2}\)+\(\dfrac{b}{c\left(c+2a\right)^2}\)+\(\dfrac{c}{a\left(a+2b\right)^2}\)≥\(\dfrac{4}{3}\)
cho 3 số thực dương thỏa mãn \(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=3\).Chứng minh rằng
\(\frac{27a^2}{c\left(c^2+9a^2\right)}+\frac{b^2}{a\left(4a^2+b^2\right)}+\frac{8c^2}{b\left(9b^2+4c^2\right)}\ge\frac{3}{2}\)
Đặt \(a=\frac{1}{x};b=\frac{2}{y};c=\frac{3}{z}\)
Theo bài ra, ta có:
x+y+z=3
\(bđt\Leftrightarrow\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2}\ge\frac{3}{2}\)
Áp dụng kĩ thuật Cau-chy ngược dấu ta có:
\(\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2}\ge\frac{x+y+z}{2}=\frac{3}{2}\)
Dấu '=' xảy ra <=> a=3;b=2;c=1
*Bài khá giống bạn kia :)
Đặt \(a=\frac{1}{x};b=\frac{2}{y};c=\frac{3}{z}\)
\(\Rightarrow x+y+z=3\)
BĐT cần chứng minh trở thành :
\(\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2}\ge\frac{3}{2}\)
Áp dụng kĩ thuật Cô Si ngược dấu ta có :
\(\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2}\ge\frac{x+y+z}{2}=\frac{3}{2}\)
Dấu đẳng thức xảy ra \(\Leftrightarrow a=3;b=2;c=1\)
Cho các số thực dương a,b,c thay đổi thỏa mãn \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\)
Tìm GTLN của P=\(\dfrac{1}{\left(2a+b+c\right)^2}+\dfrac{1}{\left(2b+c+a\right)^2}+\dfrac{1}{\left(2c+a+b\right)^2}\)
\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Tương tự và cộng lại:
\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)
\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)
Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z
\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)
Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))
\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))
\(\Rightarrow P\le\dfrac{3}{16}\)
\(ĐTXR\Leftrightarrow a=b=c=1\)
a, Giải phương trình: 2\(\left(x-\sqrt{2x^2+5x-3}\right)=1+x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)
b, Cho ba số thực dương a,b,c thỏa mãn a,b,c=1
Chứng minh rằng:\(\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2}\)
Cho các số thực a,b,c thỏa mãn a>1 , b>\(\dfrac{1}{2}\) , \(c>\dfrac{1}{3}\) và \(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\). Tìm GTLN của bt \(P=\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\)
Cho a, b, c là ba số dương thỏa mãn \(abc\)=1. Chứng minh rằng:
\(\dfrac{1}{a^3\left(b+c\right)}\)+\(\dfrac{1}{b^3\left(a+c\right)}\)+\(\dfrac{1}{c^3\left(a+b\right)}\)≥\(\dfrac{3}{2}\)
Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ca\right)^2}{b\left(c+a\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)
\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).
ĐTXR \(\Leftrightarrow a=b=c=1\)
cho a,b,c là các số thực dương. Chứng minh rằng :
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(a+b+c\right)\)
AD bđt AM-GM cho 3 số
\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+C}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c}{a^3\left(b+c\right)}.\dfrac{\left(b+c\right)}{4bc}.\dfrac{1}{2b}}=\dfrac{3}{2a}\)
\(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}\ge\dfrac{3}{2a}-\dfrac{3}{4b}-\dfrac{1}{4c}\)
thiết lập bđt tương tự r cộng lại \(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\left(\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}\right)\left(a+b+c\right)=\dfrac{1}{2}\left(a+b+c\right)\)