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Theo giả thiết, ta có: \(a^2b^2+b^2c^2+c^2a^2=a^2b^2c^2\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1\)
Áp dụng BĐT AM - GM cho 5 số, ta được: \(\hept{\begin{cases}a.a.a.b.b\le\frac{a^5+a^5+a^5+b^5+b^5}{5}=\frac{3a^5+2b^5}{5}\\b.b.b.a.a\le\frac{b^5+b^5+b^5+a^5+a^5}{5}=\frac{3b^5+2a^5}{5}\end{cases}}\)
\(\Rightarrow\frac{5\left(a^5+b^5\right)}{5}\ge a^2b^2\left(a+b\right)\)hay \(a^5+b^5\ge a^2b^2\left(a+b\right)\)
\(\Rightarrow\frac{1}{\sqrt{a^5+b^5}}\le\frac{1}{ab\sqrt{a+b}}\)(1) .
Tương tự, ta có: \(\frac{1}{\sqrt{b^5+c^5}}\le\frac{1}{bc\sqrt{b+c}}\)(2); \(\frac{1}{\sqrt{c^5+a^5}}\le\frac{1}{ca\sqrt{c+a}}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(VT=\Sigma_{cyc}\frac{1}{\sqrt{a^5+b^5}}\le\Sigma_{cyc}\frac{1}{ab\sqrt{a+b}}\)()
Xét \(\left(\Sigma_{cyc}\frac{1}{ab\sqrt{a+b}}\right)^2\le\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\left(\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}\right)\)\(=\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}\Rightarrow\Sigma_{cyc}\frac{1}{ab\sqrt{a+b}}\le\sqrt{\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}}\)(2)
Từ (1) và (2) suy ra \(\Sigma_{cyc}\frac{1}{\sqrt{a^5+b^5}}\le\sqrt{\Sigma_{cyc}\frac{1}{b^2\left(a+b\right)}}\)(đpcm)
Đẳng thức xảy ra khi \(a=b=c=\sqrt{3}\)
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
Đặt \(\left(a,b,c\right)\rightarrow\left(\frac{x}{y},\frac{y}{z},\frac{z}{x}\right)\)
\(VT=\Sigma_{cyc}\frac{1}{\sqrt{\frac{x}{z}+\frac{x}{y}+2}}=\Sigma_{cyc}\frac{\sqrt{yz}}{\sqrt{xy+xz+2yz}}\)
\(\Rightarrow VT^2\le\left(1+1+1\right)\left(\Sigma_{cyc}\frac{yz}{xy+xz+2yz}\right)\)\(\le\frac{3}{4}\left[\Sigma_{cyc}yz\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)\right]=\frac{9}{4}\)
Đẳng thức xảy ra khi a = b = c = 1
Bài 1: Bổ đề: \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
\(P=\frac{1}{\sqrt{2}}\left(\sqrt{4a^2+2ab+4b^2}+\sqrt{4b^2+2bc+4c^2}+\sqrt{4c^2+2ca+4a^2}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{3\left(a^2+b^2\right)+\left(a+b\right)^2}+\sqrt{3\left(b^2+c^2\right)+\left(b+c\right)^2}+\sqrt{3\left(c^2+a^2\right)+\left(c+a\right)^2}\right)\)
\(\ge\frac{1}{\sqrt{2}}\left(\sqrt{\frac{3}{2}\left(a+b\right)^2+\left(a+b\right)^2}+\sqrt{\frac{3}{2}\left(b+c\right)^2+\left(b+c\right)^2}+\sqrt{\frac{3}{2}\left(c+a\right)^2+\left(c+a\right)^2}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{\frac{5}{2}\left(a+b\right)^2}+\sqrt{\frac{5}{2}\left(b+c\right)^2}+\sqrt{\frac{5}{2}\left(c+a\right)^2}\right)\)
\(=\frac{1}{\sqrt{2}}.\frac{\sqrt{5}}{\sqrt{2}}+\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\)\(=\frac{\sqrt{5}}{2}.2\left(a+b+c\right)=\sqrt{5}.2020\)
Dấu "=" xảy ra khi \(a=b=c=\frac{2020}{3}\)
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
Áp dụng cosi ta có
\(\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{a}{b+c}}+\frac{b+c}{2a\sqrt[4]{2}}\ge5\sqrt[5]{\frac{1}{\sqrt[4]{2^5}}}=\frac{5}{\sqrt[4]{2}}\)
Khi đó
\(4P\ge\frac{15}{\sqrt[4]{2}}+\left(4-\frac{1}{2\sqrt[4]{2}}\right)\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)\)
Mà \(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=\left(\frac{a}{b}+\frac{b}{a}\right)+...\ge6\)
=> \(4P\ge\frac{15}{\sqrt[4]{2}}+\left(4-\frac{1}{2\sqrt[4]{2}}\right).6=24+\frac{12}{\sqrt[4]{2}}\)
=> \(P\ge6+\frac{3}{\sqrt[4]{2}}\)
dấu bằng xảy ra khi a=b=c