Các câu hỏi tương tự
11 tháng 5 2022 lúc 20:00
Cho 3 số thực dương x,y,z thỏa mãn \(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}=3\)
Chứng minh \(\dfrac{27a^2}{c\left(c^2+9a^2\right)}+\dfrac{b^2}{a\left(4a^2+b^2\right)}+\dfrac{8c^3}{b\left(9b^2+4c^2\right)}\ge\dfrac{3}{2}\)
đặt Psqrt{frac{left(a+bright)^3}{8ableft(4a+4b+cright)}}+sqrt{frac{left(b+cright)^3}{8bcleft(4b+4c+aright)}}+sqrt{frac{left(c+aright)^3}{8caleft(4c+4a+bright)}}Q8ab(4a+4b+c)+8bc(4b+4c+a)+8ca(4c+4a+b)32(a+b+c)(ab+bc+ca)-72abcáp dụng holder ta có:P^2Qge8left(a+b+cright)^3theo schur thì left(a+b+cright)^3ge4left(a+b+cright)left(ab+bc+caright)-9abcRightarrow8left(a+b+cright)^3ge32left(a+b+cright)left(ab+bc+caright)-72abcRightarrow P^2gefrac{8left(a+b+cright)^3}{Q}ge1left(Q.E.Dright)
Đọc tiếp
đặt \(P=\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}\)
Q=8ab(4a+4b+c)+8bc(4b+4c+a)+8ca(4c+4a+b)
=32(a+b+c)(ab+bc+ca)-72abc
áp dụng holder ta có:
\(P^2Q\ge8\left(a+b+c\right)^3\)
theo schur thì \(\left(a+b+c\right)^3\ge4\left(a+b+c\right)\left(ab+bc+ca\right)-9abc\)
\(\Rightarrow8\left(a+b+c\right)^3\ge32\left(a+b+c\right)\left(ab+bc+ca\right)-72abc\)
\(\Rightarrow P^2\ge\frac{8\left(a+b+c\right)^3}{Q}\ge1\left(Q.E.D\right)\)
CMR với bất kì các số thực dương a,b,c sao cho a+b+c=ab+bc+ac , bất đẳng thức sau đây xảy ra :
\(3+\sqrt[3]{\dfrac{a^3+1}{2}}+\sqrt[3]{\dfrac{b^3+1}{2}}+\sqrt[3]{\dfrac{c^3+1}{2}}\le2\left(a+b+c\right)\)
a,b,c là các số thực dương thỏa mãn a+b+c=3. CMR: \(\dfrac{a\left(a+bc\right)^2}{b\left(ab+2c^2\right)}+\dfrac{b\left(b+ca\right)^2}{c\left(bc+2a^2\right)}+\dfrac{c\left(c+ab\right)^2}{a\left(ca+2b^2\right)}>=4\)
cho a,b,c>0;\(a+b+c,abc=1\).CMR
\(\dfrac{bc}{a^2\left(b+c\right)}+\dfrac{ca}{b^2\left(c+a\right)}+\dfrac{ab}{c^2\left(a+b\right)}\ge\dfrac{3}{2}\)
Cho a,b,c>0 thỏa mãn ab+bc+ca=1. CMR:
\(\left(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\right)^3\le\dfrac{3}{2}\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\)
3 tháng 10 2021 lúc 9:15
Cho 3 số thực a,b,c thỏa mãn: \(3\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=404\)
Tìm MaxP \(=\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}+\dfrac{1}{\sqrt{5b^2+2bc+2c^2}}+\dfrac{1}{\sqrt{5c^2+2ca+2a^2}}\)
cho \(a,b,c>0\).CMR
\(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\ge\dfrac{a+b+c}{3}\)
cho a;b;c là các số thực dương thỏa mãn \(a^2+b^2+c^2=\frac{1}{3}\)CMR:\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}\ge a+b+c\)
Cho \(a;b;c\ge0\) thỏa \(a^3+b^3+c^3=3\)
Tìm GTNN của \(B=\dfrac{ab+bc+ca+a^3+b^3+c^3}{5\left(ab+bc+ca\right)+1}\)