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\(VT=\sqrt{\frac{ab+2c^2}{a^2+ab+b^2}}+\sqrt{\frac{bc+2a^2}{b^2+bc+c^2}}+\sqrt{\frac{ca+2b^2}{c^2+ca+a^2}}\)
\(=\frac{ab+2c^2}{\sqrt{\left(a^2+ab+b^2\right)\left(ab+2c^2\right)}}+\frac{bc+2a^2}{\sqrt{\left(b^2+bc+c^2\right)\left(bc+2a^2\right)}}+\frac{ca+2b^2}{\sqrt{\left(c^2+ca+a^2\right)\left(ca+2b^2\right)}}\)
\(\ge\frac{2\left(ab+2c^2\right)}{a^2+b^2+2c^2+2ab}+\frac{2\left(bc+2a^2\right)}{2a^2+b^2+c^2+2bc}+\frac{2\left(ca+2b^2\right)}{a^2+2b^2+c^2+2ca}\)
\(\ge\frac{ab+2c^2}{a^2+b^2+c^2}+\frac{bc+2a^2}{a^2+b^2+c^2}+\frac{ca+2b^2}{a^2+b^2+c^2}=ab+bc+ca+2\left(a^2+b^2+c^2\right)\)
\(=2+ab+bc+ca=VP\) (Do a2 + b2 + c2 = 1) => ĐPCM.
Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{\sqrt{3}}.\)
chăc là .............................. điền đi sẽ biếc a you ok ?
ta có \(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\frac{ab+2c^2}{\sqrt{1+ab-c^2}.\sqrt{ab+2c^2}}=\frac{ab+2c^2}{\sqrt{1+ab-c^2}\sqrt{ab+2c^2}}\)
Áp dụng bất đẳng thức cô si ta có
\(\sqrt{ab+1-c^2}\sqrt{ab+2c^2}\le\frac{1}{2}\left(ab+1-c^2+ab+2c^2\right)=\frac{1}{2}\left(2ab+1+c^2\right)\)
=\(\frac{1}{2}\left(2ab+a^2+b^2+2c^2\right)=\frac{1}{2}\left[\left(a+b\right)^2+2c^2\right]\le\frac{1}{2}\left(2a^2+2b^2+2c^2\right)=\left(a^2+b^2+c^2\right)\) =1
=> \(\frac{ab+2c^2}{...}\ge\frac{ab+2c^2}{1}=2c^2+ab\)
tương tự + vào thì e sẽ ra điều phải chứng minh
Nhà hàng Tôm hùm kính chào quý khách ĐC : 255 Nguyễn Huệ, Q tân bình , TP HCM
Ta có: \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}=\frac{a^2+ab+1}{\sqrt{a^2+ab+2ab+c^2}}\ge\frac{a^2+ab+1}{\sqrt{a^2+ab+a^2+b^2+c^2}}=\sqrt{a^2+ab+1}\)
\(\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}=\sqrt{\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2+a^2+c^2}\)
\(=\frac{1}{\sqrt{5}}.\sqrt{\left(\frac{9}{4}+\frac{3}{4}+1+1\right)\left(\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2+a^2+c^2\right)}\)
\(\ge\frac{1}{\sqrt{5}}\sqrt{\left(\frac{3}{2}\left(a+\frac{b}{2}\right)+\frac{3}{2}b+a+c\right)^2}\)
\(=\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)
=> \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)
Tương tự ta cũng chứng minh đc:
\(\frac{b^2+bc+1}{\sqrt{b^2+3bc+a^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}b+\frac{3}{2}c+a\right)\)
\(\frac{c^2+ca+1}{\sqrt{c^2+3ca+b^2}}\ge\frac{1}{\sqrt{5}}\left(\frac{5}{2}c+\frac{3}{2}a+b\right)\)
=> \(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}+\frac{b^2+bc+1}{\sqrt{b^2+3bc+a^2}}+\frac{c^2+ca+1}{\sqrt{c^3+3ca+b^2}}\ge\frac{1}{\sqrt{5}}\left(5a+5b+5c\right)\)
\(=\sqrt{5}\left(a+b+c\right)\)
Dấu "=" xảy ra <=> a = b = c =\(\frac{1}{\sqrt{3}}\)
\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)
Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)
\(P=a^2+b^2+c^2+\frac{8abc}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\)
\(=a^2+b^2+c^2+\frac{8abc}{\sqrt{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}}\)
\(=a^2+b^2+c^2+\frac{8abc}{\sqrt{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}}\)
\(=a^2+b^2+c^2+\frac{8abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Ta có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca=1\left(1\right)\)
Áp dụng BĐT Cô-si ta có:
\(a+b\ge2\sqrt{ab}\)
Tương tự:\(b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\left(2\right)\)
Từ (1) và (2) suy ra:
\(P\ge1+\frac{8abc}{8abc}=2\left(đpcm\right)\)
Dấu '=' xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
:))
ở phần cô si phần cuối là bn sai r
vì >= nhưng ở dưới mẫu nên bị đảo lại thành =< nên bn lm như thế k đúng
đay là link giải https://diendan.hocmai.vn/threads/bdt-a-2-b-2-c-2-dfrac-8abc-a-b-b-c-c-a-geq-2.341255/
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
Cosi + Svac-xơ
Có : \(3=a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(a+b+c\le3\)
\(\frac{1}{4-\sqrt{ab}}+\frac{1}{4-\sqrt{bc}}+\frac{1}{4-\sqrt{ca}}\le\frac{1}{4-\frac{a+b}{2}}+\frac{1}{4-\frac{b+c}{2}}+\frac{1}{4-\frac{c+a}{2}}\)
\(=-\left(\frac{1}{\frac{a+b}{2}-4}+\frac{1}{\frac{b+c}{2}-4}+\frac{1}{\frac{c+a}{2}-4}\right)\le\frac{-\left(1+1+1\right)^2}{a+b+c-12}=\frac{-9}{3-12}=1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
Chú ý đến giả thiết a + b + c = 1 ta viết được \(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1-c\right)\left(1+c\right)}}=\)\(\frac{ab}{\left(a+b\right)\sqrt{1-c^2}}=\frac{ab}{\left(a+b\right)\sqrt{\left(a+b+c\right)^2-c^2}}\)\(=\frac{ab}{\left(a+b\right)\sqrt{a^2+b^2+2\left(ab+bc+ca\right)}}\)
Mặt khác áp dụng bất đẳng thức Cauchy ta được \(a^2+b^2+2\left(ab+bc+ca\right)\ge2ab+2\left(ab+bc+ca\right)=\)\(2\left(ab+bc\right)+2\left(ab+ca\right)\)và \(a+b\ge2\sqrt{ab}\)
Từ đó dẫn đến \(\frac{ab}{\left(a+b\right)\sqrt{a^2+b^2+2\left(ab+bc+ca\right)}}\le\frac{ab}{2\sqrt{ab}\sqrt{2\left(ab+bc\right)+2\left(ab+ca\right)}}\)\(=\frac{1}{2}\sqrt{\frac{ab}{2\left(ab+bc\right)+2\left(ab+ca\right)}}\)
Mà theo bất đẳng thức quen thuộc \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) ta có: \(\sqrt{\frac{ab}{2\left(ab+bc\right)+2\left(ab+ca\right)}}\le\sqrt{\frac{1}{4}\left(\frac{ab}{2\left(ab+bc\right)}+\frac{ab}{2\left(ab+ca\right)}\right)}\)
\(=\frac{1}{2\sqrt{2}}\sqrt{\frac{ab}{ab+bc}+\frac{ab}{ab+ca}}=\frac{1}{2\sqrt{2}}\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}\)
Từ đó ta có bất đẳng thức: \(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}\le\frac{1}{4\sqrt{2}}\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}\)(1)
Hoàn toàn tương tự, ta có: \(\frac{bc}{\sqrt{\left(1-a\right)^3\left(1+a\right)}}\le\frac{1}{4\sqrt{2}}\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}\)(2) ; \(\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{1}{4\sqrt{2}}\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\)(3)
Cộng theo vế 3 bất đẳng thức (1), (2), (3), ta được: \(\frac{ab}{\sqrt{\left(1-c\right)^3\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^3\left(1+c\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\)\(\le\frac{1}{4\sqrt{2}}\left(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\right)\)
Ta cần chứng minh\(\frac{1}{4\sqrt{2}}\left(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\right)\le\frac{3\sqrt{2}}{8}\)
Hay \(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\le3\)
Áp dụng bất đẳng thức Bunhiacopxki ta được \(\sqrt{\frac{a}{a+c}+\frac{b}{b+c}}+\sqrt{\frac{b}{b+a}+\frac{c}{c+a}}+\sqrt{\frac{c}{c+b}+\frac{a}{a+b}}\)
\(\le\sqrt{3\left(\frac{a}{a+c}+\frac{b}{b+c}+\frac{b}{b+a}+\frac{c}{c+a}+\frac{c}{c+b}+\frac{a}{a+b}\right)}=3\)
Vậy bất đẳng thức được chứng minh
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Sửa đề: \(\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\)