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1. Ta có : \(\frac{a}{a+b+c+d}< \frac{a}{a+b+c}< \frac{a+d}{a+b+c+d}\)
\(\frac{b}{a+b+c+d}< \frac{b}{b+c+d}< \frac{a+b}{a+b+c+d}\)
\(\frac{c}{a+b+c+d}< \frac{c}{a+c+d}< \frac{b+c}{a+b+c+d}\)
\(\frac{d}{a+b+c+d}< \frac{d}{a+b+d}< \frac{c+d}{a+b+c+d}\)
Cộng vế theo vế ta được :
\(1< \frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2\) ( đpcm )
2. Áp dụng bất đẳng thức Cô - si cho 2 số ko âm b-1 và 1 ta có :
\(\sqrt{\left(b-1\right)\cdot1}\le\frac{\left(b-1\right)+1}{2}=\frac{b}{2}\)
Dấu "=" xảy ra <=> b - 1 = 1 <=> b = 2
\(\Rightarrow a\sqrt{b-1}=a\sqrt{\left(b-1\right)\cdot1}\le a\cdot\frac{b}{2}=\frac{ab}{2}\)
Tương tự ta có : \(b\sqrt{a-1}\le\frac{ab}{2}\) Dấu "=" xảy ra <=> a = 2
Do đó : \(a\sqrt{b-1}+b\sqrt{a-1}\le\frac{ab}{2}+\frac{ab}{2}=ab\)
Dấu "=" xảy ra <=> a = b = 2
Ta có BĐT cần chứng minh tương đương với:
\(\frac{a}{2}-\frac{a^2}{2a+1}+\frac{b}{2}-\frac{b^2}{2b+1}+\frac{c}{2}-\frac{c^2}{2c+1}\ge\frac{a+b+c}{2}-\frac{a^2+b^2+c^2}{\sqrt{a^2+b^2+c^2+6}}\)
Hay: \(\frac{a}{2a+1}+\frac{b}{2b+1}+\frac{c}{2c+1}+\frac{2\left(a^2+b^2+c^2\right)}{\sqrt{a^2+b^2+c^2+6}}\ge3\)
Áp dụng BĐT Bunhiacopxki dạng dạng p.thức ta được:
\(\frac{a}{2a+1}+\frac{b}{2b+1}+\frac{c}{2c+1}\ge\frac{\left(a+b+c\right)^2}{2\left(a^2+b^2+c^2\right)+3}\)
Khi đó ta cần chứng minh:
\(\frac{9}{2\left(a^2+b^2+c^2\right)+3}+\frac{2\left(a^2+b^2+c^2\right)}{\sqrt{a^2+b^2+c^2+6}}\ge3\)
Đặt: \(t=a^2+b^2+c^2\ge3\) ta có:
\(\frac{9}{2t+3}+\frac{2t}{\sqrt{t+6}}\ge3\Leftrightarrow\frac{9}{2t+3}-1+\frac{2t}{\sqrt{t+6}}-2\ge0\)
\(\Leftrightarrow\frac{2\left(3-t\right)}{2t+3}+\frac{2t-2\sqrt{t+6}}{\sqrt{t+6}}\ge0\)
\(\Leftrightarrow\left(t-3\right)\left[\frac{t+2}{\sqrt{t+6}\left(t+\sqrt{t+6}\right)}-\frac{1}{2t+3}\right]\ge0\)
\(\Leftrightarrow\left(t+2\right)\left(2t+3\right)-\sqrt{t+6}\left(t+\sqrt{t+6}\right)\ge0\)
\(\Leftrightarrow t\left(2t+6-\sqrt{t+6}\right)\ge0\)
Vì: \(t\ge3\) nên BĐT luôn đúng.
BĐT xảy ra \(\Leftrightarrow a=b=c=1\)
Sử dụng Bunhiacopxki:
\(\sqrt{\left(\Sigma_{cyc}\frac{a^2}{\sqrt{a^2+b^2+c^2+6}}\right)\left(\Sigma_{cyc}\frac{a^2\sqrt{a^2+b^2+c^2+6}}{\left(2a+1\right)^2}\right)}\ge\Sigma_{cyc}\frac{a^2}{2a+1}=VT\)
Hay: \(\sqrt{VP.\left(\Sigma_{cyc}\frac{a^2\sqrt{a^2+b^2+c^2+6}}{\left(2a+1\right)^2}\right)}\ge VT\)
Vậy ta chỉ cần chứng minh: \(VP\ge\sqrt{VP.\left(\Sigma_{cyc}\frac{a^2\sqrt{a^2+b^2+c^2+6}}{\left(2a+1\right)^2}\right)}\)
\(\Leftrightarrow VP\ge\Sigma_{cyc}\frac{a^2\sqrt{a^2+b^2+c^2+6}}{\left(2a+1\right)^2}\)
\(\Leftrightarrow\frac{a^2+b^2+c^2}{a^2+b^2+c^2+6}\ge\Sigma_{cyc}\frac{a^2}{\left(2a+1\right)^2}\)
Ta có
\(\sqrt{\frac{ab}{c+ab}}=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}\)\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\)\(=\sqrt{\frac{a}{c+a}}.\sqrt{\frac{b}{c+b}}\)\(\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
Tương tự, ta có
\(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\)
\(\sqrt{\frac{ca}{b+ca}\le\frac{1}{2}\left(\frac{c}{c+b}+\frac{a}{b+a}\right)}\)
Cộng vế theo vế của 3 bđt ta được đpcm
Ta có:\(a^5+ab+b^2\ge3a^2b\)
Tương tự ta có:
\(VT\le\frac{1}{\sqrt{3ab\left(a+2c\right)}}+\frac{1}{\sqrt{3bc\left(b+2a\right)}}+\frac{1}{\sqrt{3ca\left(c+2b\right)}}\)
\(=\frac{1}{\sqrt{3}}\left(\sqrt{\frac{c}{c+2a}}+\sqrt{\frac{a}{b+2a}}+\sqrt{\frac{b}{2b+c}}\right)\)
Ta cũng có:\(a+2c=a+c+c\ge\frac{1}{3}\left(\sqrt{a}+2\sqrt{c}\right)^2\)
\(\Rightarrow VT\le\frac{\sqrt{c}}{\sqrt{a}+2\sqrt{c}}+\frac{\sqrt{a}}{\sqrt{b}+2\sqrt{a}}+\frac{\sqrt{b}}{\sqrt{c}+2\sqrt{b}}\)
Đặt \(x=\frac{\sqrt{a}}{\sqrt{c}};y=\frac{\sqrt{b}}{\sqrt{a}};z=\frac{\sqrt{c}}{\sqrt{b}};xyz=1\)
\(\Rightarrow VT\le\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\)
Giả sử \(xy\le1\) thì \(z\ge1\)
Ta có: \(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}=\frac{1}{2}\left(\frac{1}{\frac{x}{2}+1}+\frac{1}{\frac{y}{2}+1}\right)+\frac{1}{z+2}\)
\(\le\frac{1}{1\frac{\sqrt{xy}}{2}}+\frac{1}{z+2}\le1\)(Đpcm)
Dấu = khi \(a=b=c=1\)
Ta có: \(\left(a^4-a^3+2\right)-\left(a+1\right)=\left(a-1\right)^2\left(a^2+a+1\right)\ge0\)\(\Rightarrow a^4-a^3+2\ge a+1\Leftrightarrow a^4-a^3+ab+2\ge ab+a+1\)
\(\Rightarrow\frac{1}{\sqrt{a^4-a^3+ab+2}}\le\frac{1}{\sqrt{ab+a+1}}\)
Tương tự:\(\frac{1}{\sqrt{b^4-b^3+bc+2}}\le\frac{1}{\sqrt{bc+b+1}}\); \(\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\frac{1}{\sqrt{ca+c+1}}\)
\(\Rightarrow VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\)\(\le\sqrt{3\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)}\)\(\le\sqrt{3\left(\frac{c}{abc+ac+c}+\frac{ac}{abc^2+abc+ac}+\frac{1}{ca+c+1}\right)}\)\(\le\sqrt{3\left(\frac{c}{ac+c+1}+\frac{ac}{ac+c+1}+\frac{1}{ca+c+1}\right)}=\sqrt{3}\)(abc = 1)
Đẳng thức xảy ra khi a = b = c = 1