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Ta có: \(4b\sqrt{c}-c\sqrt{a}=\sqrt{c}\left(4b-\sqrt{ac}\right)>0\)( do \(1< a,b,c< 2\))
Tương tự => Các MS dương
\(VT=\frac{ba}{4b\sqrt{ac}-ca}+\frac{cb}{4c\sqrt{ba}-ab}+\frac{ac}{4a\sqrt{bc}-bc}\)
Áp dụng BĐT cosi schawr ta có
\(VT\ge\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}{4b\sqrt{ac}+4c\sqrt{ab}+4a\sqrt{bc}-ab-bc-ac}\)
Áp dụng cosi ta có \(2b\sqrt{ac}=2\sqrt{ab}.\sqrt{ac}\le ab+ac\);\(2c\sqrt{ab}\le ac+bc\);\(2a\sqrt{bc}\le ab+ac\)
=> \(VT\ge\frac{\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2}{ab+bc+ac+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}}=\frac{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}{\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)^2}=1\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c
đặt \(S=\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\)
\(=\frac{a^3}{4a^2b^2+a^2}+\frac{b^3}{4b^2c^2+b^2}+\frac{c^3}{4a^2c^2+c^2}\ge\frac{\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2}{4a^2b^2+4b^2c^2+4c^2a^2+a^2+b^2+c^2}\)
xét hiệu:
1-4(a2b2+b2c2+c2a2)-a2-b2-c2
=2ab+2bc+2ca-4(a2b2+b2c2+c2a2)
=2ab(1-2ab)+2bc(1-2bc)+2ca(1-2ca)
ta có:
\(2ab\le\frac{\left(a+b\right)^2}{2}\le\frac{1}{2};2bc\le\frac{\left(b+c\right)^2}{2}\le\frac{1}{2};2ca\le\frac{\left(c+a\right)^2}{2}\le\frac{1}{2}\)
\(\Rightarrow2ab\left(1-2ab\right);2bc\left(1-2bc\right);2ca\left(1-2ca\right)\ge0\)
\(\Rightarrow1\ge4\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2\)
\(\Rightarrow\frac{\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2}{4\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2}\ge\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2\)
\(\Rightarrow\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\ge\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2\)
=>đpcm
dấu"=" xảy ra khi 1 số=1;2 số còn lại =0
Thôi giải lại câu 1:v (ý tưởng dồn biến là quá trâu bò! Bên AoPS em mới phát hiện ra có một cách bằng Cauchy-Schwarz quá hay!)
\(BĐT\Leftrightarrow\Sigma_{cyc}\frac{\left(a+b+c\right)^2}{2a^2+\left(a^2+b^2\right)+\left(a^2+c^2\right)}\le\frac{9}{2}\)(*)
BĐT này đúng theo Cauchy-Schwarz: \(VT_{\text{(*)}}\le\Sigma_{cyc}\left(\frac{a^2}{2a^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)=\frac{9}{2}\)
Ta có đpcm.
Equality holds when a = b = c = 1 (Đẳng thức xảy ra khi a = b =c = 1)
Ap dung BDT Bun-hia-cop-xki ta co:
\(\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\le\left(1+1+1\right)\left[4\left(a+b+c\right)+3\right]=21\)
\(\Rightarrow-\sqrt{21}\le\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{21}< 5\)
\(\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 5\)
Ap dung BDT Bun-hia-cop-xki ta co:
\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\le\left(1+1+1\right)\left[4\left(a+b+c\right)+3\right]=21(4a+1+4b+1+4c+1)2≤(1+1+1)[4(a+b+c)+3]=21
\Rightarrow-\sqrt{21}\le\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{21}< 5⇒−21≤4a+1+4b+1+4c+1≤21<5
\Rightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}< 5⇒4a+1+4b+1+4c+1<5
2) \(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+\dfrac{1}{4xy}\)
Áp dụng BĐT Cauchy-Schwa, ta có:
\(A\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x+y\right)^2}=\dfrac{3}{2}\)
1) Áp dụng BĐT Bunyakovsky, ta có:
\(\left(4a+1+4b+1+4c+1\right)3\ge\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\)
\(\Rightarrow VT\le\sqrt{21}< 3\)(Sai)
Vậy đề sai, thử với a=0,5;b=0,1;c=0,4
áp dụng bất đẳng thức: (a+b+c)^2<=3(a^2+b^2+c^2):
[√(4a+1)+√(4b+1)+√(4c+1)]^2
<= 3[4(a+b+c)+3]=21<25
=>√(4a+1)+√(4b+1)+√(4c+1)<5
cosi : \(\sqrt{4a+1}\)\(\sqrt{1}\)<\(\frac{4a+1+1}{2}\)= 2a + 1. tương tự \(\sqrt{4b+1}\)\(\sqrt{1}\)<\(\frac{4b+1+1}{2}\)= 2b + 1; \(\sqrt{4c+1}\)\(\sqrt{1}\)<\(\frac{4c+1+1}{2}\)= 2c + 1. Nên VT < 2(a+b+c) +3 = 5. Dấu = xảy ra khi và chỉ khi a=b=c = 1/3
\(A\le\frac{1}{27}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^3\)
Mặt khác :
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{3\left[4\left(a+b+c\right)+3\right]}\)
\(=3\sqrt{5}\)
\(\Rightarrow A\le\frac{1}{27}\left(3\sqrt{5}\right)^3=5\sqrt{5}\)
Dấu " = " xảy ra khi \(a=b=c=1\)
Ta có
\(\sqrt{2}\sqrt{4a+1}\le\frac{4a+3}{2}\)
\(\sqrt{2}\sqrt{4b+1}\le\frac{4b+3}{2}\)
\(\sqrt{2}\sqrt{4c+1}\le\frac{4c+3}{2}\)
\(\sqrt{2}\sqrt{4d+1}\le\frac{4d+3}{2}\)
Cộng vế theo vế ta được
\(\sqrt{2}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}\right)\)
\(\le8\)
<=> \(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\sqrt{4d+1}\le4\sqrt{2}\)