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1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)
\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)
\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)
\(=\frac{1}{\left(a+b+c\right)a+bc}+\frac{1}{\left(a+b+c\right)b+ac}+\frac{1}{\left(a+b+c\right)c+ab}\)
\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
Vậy VT = VP, đẳng thức được chứng minh
Áp dụng bất đẳng thức Cô-si :
\(a\sqrt{3a\left(a+2b\right)}+b\sqrt{3b\left(b+2a\right)}\le a\cdot\frac{3a+a+2b}{2}+b\cdot\frac{3b+b+2a}{2}\)
\(=a\cdot\frac{4a+2b}{2}+b\cdot\frac{4b+2a}{2}\)
\(=a\left(2a+b\right)+b\left(2b+a\right)\)
\(=2a^2+2b^2+2ab\)
\(=2\left(a^2+b^2+ab\right)\le2\left(2+\frac{a^2+b^2}{2}\right)=2\left(2+\frac{2}{2}\right)=6\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=1\)
p/s: có gì chiều giải nốt, giờ đi ăn cơm @@
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
a) Ta có: \(a^2-1\le0;b^2-1\le0;c^2-1\le0\)
\(\Rightarrow\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\le0\)
\(a^2+b^2+c^2\le1+a^2b^2+b^2c^2+c^2a^2-a^2b^2c^2\le1+a^2b^2+b^2c^2+c^2a^2\) ( vì \(abc\ge0\) )
Có \(b-1\le0\Rightarrow a^2b\sqrt{b}\left(b-1\right)\le0\Rightarrow a^2b^2\le a^2b\sqrt{b}\)
Tương tự: \(\hept{\begin{cases}b^2c^2\le b^2c\sqrt{c}\\c^2a^2\le c^2a\sqrt{a}\end{cases}\Rightarrow dpcm}\)
Bài 1, t nghĩ VP căn phải kéo dài hết
Áp dụng bđt bu nhi a, ta có
\(\left(\sqrt{ab}+\sqrt{cd}\right)^2\le\left(a+d\right)\left(b+c\right)\Rightarrow\sqrt{ab}+\sqrt{cd}\le\sqrt{\left(a+d\right)\left(b+c\right)}\left(ĐPCM\right)\)
Bài 2, Áp dụng bài 1, ta có
\(\left(a\sqrt{3a\left(a+2b\right)}+b\sqrt{3b\left(b+2a\right)}\right)\le\left(a^2+b^2\right)\left[3a\left(a+2b\right)+3b\left(b+2a\right)\right]\)
\(\le2\left(3a^2+6ab+3b^2+6ab\right)=2\left[3\left(a^2+b^2\right)+12ab\right]\le2\left(6+12ab\right)\)
Áp dụng bđt cô si, ta có
\(a^2+b^2\ge2ab\Rightarrow2\ge2ab\Rightarrow12\ge12ab\)
=>(...)^2<=36 => ...<=6 (ĐPcM)
dấu = xảy ra <=> a=b=1
^_^