Cho a;b;c>0.CMR:
\(\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}+\sqrt[3]{\frac{b^2+ca}{abc\left(c^2+a^2\right)}}+\sqrt[3]{\frac{c^2+ab}{abc\left(a^2+b^2\right)}}\ge\frac{9}{a+b+c}\)
+) 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: \(\dfrac{2017a-a^2}{bc}+\dfrac{2017b-b^2}{ca}+\dfrac{2017c-c^2}{ab}\ge\sqrt{2}\left(\Sigma\sqrt{\dfrac{2017-a}{a}}\right)\)
1. a,b,c>0 và a+b+c=2017
\(CM:\Sigma\dfrac{2017a-a^2}{bc}\ge\sqrt{2}\left(\Sigma\sqrt{\dfrac{2017-a}{a}}\right)\)
2. cho x,y,z tm: \(x^2+y^2+z^2=3\)
\(CM:8\left(2-x\right)\left(2-y\right)\left(2-z\right)\ge\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)\)
3. a,b,c>0 và \(a^2+b^2+c^2\ge6\)
\(CM:\Sigma\dfrac{1}{1+ab}\ge\dfrac{3}{2}\)
Cho 3 sô dương a,b,c . Chứng mình rằng
\(\sqrt[3]{\frac{\left(a\text{+}b\right)\left(b\text{+}c\right)\left(c\text{+}a\right)}{abc}}\ge\frac{4}{3}\left(\frac{a^2}{a^2\text{+}bc}\frac{b^2}{b^2\text{+}ab}\frac{c^2}{c^2\text{+}ac}\right)\)
Mấy bạn giúp mình câu này với ;-;
Cho a, b, c > 0. Chứng minh rằng: \(\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ca}\le\dfrac{2}{3}\left(a+b+c\right)\)
Cho a; b; c > 0 sao cho a+b+c=3. Chứng minh rằng
\(\frac{a}{b^2\left(ca+1\right)}+\frac{b}{c^2\left(ab+1\right)}+\frac{c}{a^2\left(bc+1\right)}\ge\frac{9}{\left(1+abc\right)\left(ab+bc+ca\right)}\)
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 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 dương. Chứng minh rằng:
\(\sqrt[4]{\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4}-\sqrt[4]{3}\ge\dfrac{\sqrt[4]{243}}{2+abc}\)