BĐT này không đúng
Ví dụ: với \(a=b=c=0,1\)
Cho a,b,c là các số thực dương bất kì, chứng minh rằng:
\(\dfrac{\sqrt{bc}}{a+3\sqrt{bc}}+\dfrac{\sqrt{ca}}{b+3\sqrt{ca}}+\dfrac{\sqrt{ab}}{c+3\sqrt{ab}}\le\dfrac{3}{4}\)
Cho a, b, c dương. Chứng minh rằng:
\(\sqrt[4]{\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4}-\sqrt[4]{3}\ge\dfrac{\sqrt[4]{243}}{2+abc}\)
+) Giải hệ pt: \(\left\{{}\begin{matrix}4\sqrt{x^2+4y-5}=y^2-x+10\\x^3+\left(1-y\right)x^2=\left(x+4\right)y\end{matrix}\right.\)
+) Cho a,b,c>0 và a+b+c=2017
CM: \(\dfrac{2017a-a^2}{bc}+\dfrac{2017b-b^2}{ca}+\dfrac{2017c-c^2}{ab}\ge\sqrt{2}\left(\Sigma\sqrt{\dfrac{2017-a}{a}}\right)\)
cho a,b,c thực dương thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le16\left(a+b+c\right)\)
CMR:
\(\dfrac{1}{\left(a+b+2\sqrt{a+c}\right)^3}+\dfrac{1}{\left(b+c+2\sqrt{b+a}\right)^3}+\dfrac{1}{\left(c+a+2\sqrt{c+b}\right)^3}\le\dfrac{8}{9}\)
Cho a;b;c>0.CMR:
\(\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}+\sqrt[3]{\frac{b^2+ca}{abc\left(c^2+a^2\right)}}+\sqrt[3]{\frac{c^2+ab}{abc\left(a^2+b^2\right)}}\ge\frac{9}{a+b+c}\)
Cho \(a;b;c\ge0.\) Cm:
1) \(a^3+b^3+c^3+3abc\ge ab\sqrt{2\left(a^2+b^2\right)}+bc\sqrt{2\left(b^2+c^2\right)}+ca\sqrt{2\left(c^2+a^2\right)}\)
2) \(a^2+b^2+c^2+ab+bc+ca\ge a\sqrt{2\left(b^2+c^2\right)}+b\sqrt{2\left(c^2+a^2\right)}+c\sqrt{2\left(a^2+b^2\right)}\)
cho 3 số thực dương a,b,c. chứng minh
\(ab+bc+ca\le\frac{a^3\left(b+c\right)}{a^2+bc}+\frac{b^3\left(c+a\right)}{b^2+ca}+\frac{c^3\left(a+b\right)}{c^2+ab}\le a^2+b^2+c^2\)\(ab+bc+ca\le\frac{a^3\left(b+c\right)}{a^2+bc}+\frac{b^3\left(c+a\right)}{b^2+ca}+\frac{c^3\left(a+b\right)}{c^2+ab}\le a^2+b^2+c^2\)
Cho a, b, c thỏa mãn ab+bc+ca=3 CMR
\(\sqrt[3]{\frac{a}{b\left(b+2c\right)}}+\sqrt[3]{\frac{b}{c\left(c+2a\right)}}+\sqrt[3]{\frac{c}{a\left(a+2b\right)}}\ge\frac{3}{\sqrt[3]{3}}\)
Cho \(a,b,c>0\)và \(ab+bc+ca=1\). Chứng minh \(M=\frac{1-a^2}{1+a^2}+\sqrt{3}\left(a^2+b^2+c^2\right)\le\frac{1}{8}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2}+\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
