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\(\sqrt{3}>\frac{m}{n}\Rightarrow3>\frac{m^2}{n^2}\Rightarrow3n^2>m^2\Rightarrow3n^2\ge m^2+1\)
với 3n2=m2+1=>m2+1 chia hết cho 3
=>m2 chia 3 dư 2(vô lí)
\(\Rightarrow3n^2\ge m^2+2\)
lại có:\(\left(m+\frac{1}{2m}\right)^2=m^2+1+\frac{1}{4m^2}< m^2+2\)
\(\Rightarrow\left(m+\frac{1}{2m}\right)^2< 3n^2\Rightarrow m+\frac{1}{2m}< \sqrt{3}n\)
\(\Rightarrow\frac{m}{n}+\frac{1}{2mn}< \sqrt{3}\left(Q.E.D\right)\)
Khi các căn thức đều xác định, áp dụng BĐT Bunhia:
\(a\sqrt{1-b^2}+b\sqrt{1-c^2}+c\sqrt{1-a^2}\le\sqrt{\left(a^2+b^2+c^2\right)\left(3-\left(a^2+b^2+c^2\right)\right)}\)
\(\le\frac{a^2+b^2+c^2+3-\left(a^2+b^2+c^2\right)}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi
\(a^2+b^2+c^2=3-\left(a^2+b^2+c^2\right)\Leftrightarrow a^2+b^2+c^2=\frac{3}{2}\) (đpcm)
\(VT=\sum\frac{x}{x+\sqrt{\left(xy+xz+yz\right)x+yz}}=\sum\frac{x}{x+\sqrt{\left(x+y\right)\left(x+z\right)}}=\sum\frac{x}{x+\sqrt{\left(\sqrt{x}^2+\sqrt{y}^2\right)\left(\sqrt{z}^2+\sqrt{x}^2\right)}}\)
\(\Rightarrow VT\le\sum\frac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{yz}\right)^2}}=\sum\frac{x}{x+\sqrt{xz}+\sqrt{yz}}=\sum\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
\(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{5a^2+2ab+2b^2}}\le\dfrac{1}{\sqrt{\left(2a+b\right)^2}}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự ta có: \(\dfrac{1}{\sqrt{5b^2+2bc+2c^2}}\le\dfrac{1}{9}\left(\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\dfrac{1}{\sqrt{5c^2+2ac+a^2}}\le\dfrac{1}{9}\left(\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế với vế:
\(\dfrac{1}{\sqrt{5a^2+2ab+b^2}}+\dfrac{1}{\sqrt{5b^2+2bc+c^2}}+\dfrac{1}{\sqrt{5c^2+2ac+a^2}}\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\le\dfrac{2}{3}\)
Dấu "=" khi \(a=b=c=\dfrac{3}{2}\)
\(\sqrt{3}-\dfrac{m}{n}>0\Leftrightarrow\sqrt{3}>\dfrac{m}{n}\Leftrightarrow3n^2>m^2\)
Vì \(m,n\ge1\) nên \(3n^2\ge m^2+1\)
Với \(3n^2=m^2+1\Leftrightarrow m^2+1⋮3\Leftrightarrow m^2\) chia 3 dư 2 (vô lí)
\(\Leftrightarrow3n^2\ge m^2+2\)
Lại có \(4m^2>1\Leftrightarrow\left(m+\dfrac{1}{2m}\right)^2=m^2+1+\dfrac{1}{4m^2}< m^2+2\)
\(\Leftrightarrow\left(m+\dfrac{1}{2m}\right)^2< 3n^2\Leftrightarrow m+\dfrac{1}{2m}< n\sqrt{3}\\ \Leftrightarrow n\sqrt{3}-m>\dfrac{1}{2m}\)
