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Đặt vế trái là T, ta có:
\(\dfrac{a}{\sqrt{b+1}}=\dfrac{a\sqrt{2}}{\sqrt{2}.\sqrt{b+1}}\ge\dfrac{a\sqrt{2}}{\dfrac{b+1+2}{2}}=\dfrac{a.2\sqrt{2}}{b+3}\)
Tương tự: \(\dfrac{b}{\sqrt{c+1}}\ge\dfrac{b.2\sqrt{2}}{c+3}\)
\(\dfrac{c}{\sqrt{a+1}}\ge\dfrac{c.2\sqrt{2}}{a+3}\)
Cộng vế theo vế các BĐT vừa chứng minh, ta được
\(T\ge2\sqrt{2}\left(\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{a+3}\right)=2\sqrt{2}\left(\dfrac{a^2}{ab+3a}+\dfrac{b^2}{bc+3b}+\dfrac{c^2}{ac+3c}\right)\)
\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+3\left(a+b+c\right)}\)
\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{\dfrac{\left(a+b+c\right)^2}{3}+3\left(a+b+c\right)}\)
\(T\ge2\sqrt{2}.\dfrac{3^2}{\dfrac{3^2}{3}+9}=\dfrac{3\sqrt{2}}{2}\)(đpcm)
Đẳng thức xảy ra khi a=b=c=1
b) Đặt vế trái là N,ta có:
\(\sum\sqrt{\dfrac{a^3}{b+3}}=\sum\sqrt{\dfrac{a^4}{ab+3}}=\sum\dfrac{a^2}{\sqrt{ab+3}}=\sum\dfrac{2a^2}{\sqrt{4a\left(b+3\right)}}\ge\sum\dfrac{2a^2}{\dfrac{4a+b+3}{2}}=\sum\dfrac{4a^2}{4a+b+3}\)
\(\sum\dfrac{4a^2}{4a+b+3}\ge\dfrac{\left(2a+2b+2c\right)^2}{4a+b+3+4b+c+3+4c+a+3}=\dfrac{3}{2}\)(đpcm)
Đẳng thức xảy ra khi a=b=c=1
Nice proof, nhưng đã quy đồng là phải thế này :v
\(BDT\Leftrightarrow\left(2a-\sqrt{a^2+3}\right)+\left(2b-\sqrt{b^2+3}\right)+\left(2c-\sqrt{c^2+3}\right)\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}\ge0\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{a}-a\right)+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{b}-b\right)+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{c}-c\right)\ge0\)
\(\Leftrightarrow\left(a^2-1\right)\left(\dfrac{1}{2a+\sqrt{a^2+3}}-\dfrac{1}{4a}\right)+\left(b^2-1\right)\left(\dfrac{1}{2b+\sqrt{b^2+3}}-\dfrac{1}{4b}\right)+\left(c^2-1\right)\left(\dfrac{1}{2c+\sqrt{a^2+3}}-\dfrac{1}{4c}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)\left(2a-\sqrt{a^2+3}\right)}{a\left(2a+\sqrt{a^2+3}\right)}+\dfrac{\left(b^2-1\right)\left(2b-\sqrt{b^2+3}\right)}{b\left(2b+\sqrt{b^2+3}\right)}+\dfrac{\left(c^2-1\right)\left(2c-\sqrt{c^2+3}\right)}{c\left(2c+\sqrt{c^2+3}\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)^2}{a\left(2a+\sqrt{a^2+3}\right)^2}+\dfrac{\left(b^2-1\right)^2}{b\left(2b+\sqrt{b^2+3}\right)^2}+\dfrac{\left(c^2-1\right)^2}{c\left(2c+\sqrt{c^2+3}\right)^2}\ge0\) (luôn đúng)
Khi \(f\left(t\right)=\sqrt{1+t}\) là hàm lõm trên \([-1, +\infty)\) ta có:
\(f(t)\le f(3)+f'(3)(t-3)\forall t\ge -1\)
Tức là \(f\left(t\right)\le2+\dfrac{1}{4}\left(t-3\right)=\dfrac{5}{4}+\dfrac{1}{4}t\forall t\ge-1\)
Áp dụng BĐT này ta có:
\(\sqrt{a^2+3}=a\sqrt{1+\dfrac{3}{a^2}}\le a\left(\dfrac{5}{4}+\dfrac{1}{4}\cdot\dfrac{3}{a^2}\right)=\dfrac{5}{4}a+\dfrac{3}{4}\cdot\dfrac{1}{a}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\sqrt{b^2+3}\le\dfrac{5}{4}b+\dfrac{3}{4}\cdot\dfrac{1}{b};\sqrt{c^2+3}\le\dfrac{5}{4}c+\dfrac{3}{4}\cdot\dfrac{1}{c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VP\le\dfrac{5}{4}\left(a+b+c\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2\left(a+b+c\right)=VT\)
Do 1/b+1/c=3/4-1/a suy ra \(\sum\) (1a/)=3/4
Ta có \(\dfrac{\sqrt{b^2+bc+c^2}}{a^2}\)= \(\dfrac{\sqrt{\left(b+c\right)^2-bc}}{a^2}\ge\dfrac{\sqrt{\left(b+c\right)^2-\dfrac{\left(b+c\right)^2}{4}}}{a^2}=\dfrac{\sqrt{3}\left(b+c\right)}{2a^2}\)
Tương tự ta được:
P\(\ge\) \(\sqrt{3}\) \(\left(\sum\dfrac{b+c}{a^2}\right)\) \(\ge\) \(\sqrt{3}\) (1/a+1/b+1/c) \(\ge\dfrac{3\sqrt{3}}{4}\)
Đẳng thức xảy ra \(\Leftrightarrow\) a=b=c=4
Lời giải:
Từ điều kiện
\(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Rightarrow abc(a+b+c)=ab+bc+ac\)
Sử dụng hệ quả của BĐT AM-GM:
\((ab+bc+ac)^2\geq 3abc(a+b+c)\Rightarrow \frac{(ab+bc+ac)^2}{3}\geq ab+bc+ac\)
Suy ra \(ab+bc+ac\geq 3\). Do đó:
\(\text{VP}\leq \sqrt{a^2+ab+bc+ac}+\sqrt{b^2+ab+bc+ac}+\sqrt{c^2+ab+bc+ac}\)
\(\Leftrightarrow \text{VP}\leq \sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)}\)
Áp dụng BĐT AM-GM: \(\sqrt{(a+b)(a+c)}\leq \frac{a+b+a+c}{2}\) và tương tự....
\(\Rightarrow \text{VP}\leq \frac{a+b+a+c}{2}+\frac{b+c+b+a}{2}+\frac{c+a+c+b}{2}=2(a+b+c)=\text{VT}\)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=1\)
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
a) BĐT cần cm tương đương ;
\(a-\dfrac{ab^2}{1+b^2}+b-\dfrac{bc^2}{1+c^2}+a-\dfrac{a^2c}{1+a^2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow3-\left(\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\right)\ge\dfrac{3}{2}\)
\(\Leftrightarrow\left(\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\right)\le\dfrac{3}{2}\)
Áp dụng BĐT Cauchy
\(\Rightarrow\dfrac{ab^2}{1+b^2}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\)
tương tự rồi cộng vế theo vế các BĐT lại
\(\Leftrightarrow\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\le\dfrac{ab+bc+ac}{2}\)
mặt khác \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\)
\(\Rightarrow\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\le\dfrac{3}{2}\)
ĐPCM
Áp dụng BĐT AM-GM và Cauchy-Schwarz ta có:
\(VT=\dfrac{a^2}{a+abc}+\dfrac{b^2}{b+abc}+\dfrac{c^2}{c+abc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3abc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\dfrac{\left(a+b+c\right)\left(ab+bc+ca\right)}{3}}=\dfrac{3\left(a+b+c\right)}{3+ab+bc+ca}\)
Tức cần chứng minh \(\dfrac{3\left(a+b+c\right)}{3+ab+bc+ca}\ge1\)
\(\Leftrightarrow3\left(a+b+c\right)\ge3+ab+bc+ca\)
\(\Leftrightarrow9\left(a+b+c\right)^2\left(a^2+b^2+c^2\right)\ge\left(3\left(a^2+b^2+c^2\right)+ab+bc+ca\right)^2\)
Đặt \(a^2+b^2+c^2=k\left(ab+bc+ca\right)\left(k\ge1\right)\) và ta cần cm:
\(9(k+2)k\geq(3k+1)^2\)\(\Leftrightarrow12k-1\ge9\) *đúng với \(k\ge 1\) :|*
Từ \(a^2+b^2+c^2=3\Rightarrow a+b+c\le3\)
Ta có: \(\sqrt{\dfrac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\dfrac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\dfrac{9}{\left(c+a\right)^2}+b^2}\)
\(\ge\sqrt{9\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)^2+\left(a+b+c\right)^2}\)
\(\ge\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)
Cần chứng minh \(\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\ge\dfrac{3\sqrt{13}}{2}\)
\(\Leftrightarrow9\left(\dfrac{9}{2t}\right)^2+t^2\ge\dfrac{117}{4}\left(t=a+b+c\le3\right)\)
\(\Leftrightarrow\dfrac{\left(t-3\right)\left(2t-9\right)\left(t+3\right)\left(2t+9\right)}{4t^2}\ge0\)*Đúng*
B1:a)ĐK: \(x\ne 0;4;9\)
b)\(P=\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1+1}{\sqrt{x}+1}\right)\)
\(=\dfrac{x-9-x+4+x^{\dfrac{1}{2}}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{x^{\dfrac{1}{2}}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)\(=\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}\)
c)Vì \(x^{\dfrac{1}{2}}+1>0\forall x\) nên
\(P< 0< =>x-2x^{\dfrac{1}{2}}< 0\)
\(\Leftrightarrow x^{\dfrac{1}{2}}\left(x^{\dfrac{1}{2}}-2\right)< 0\)
\(\Leftrightarrow0< x< 4\)
Vậy 0<x<4 thì P<0
d)tA CÓ: \(\dfrac{1}{P}=\dfrac{x-2x^{\dfrac{1}{2}}}{x^{\dfrac{1}{2}}+1}=\dfrac{x-2x^{\dfrac{1}{2}}+1-1}{x^{\dfrac{1}{2}}+1}=\dfrac{\left(x^{\dfrac{1}{2}}-1\right)^2-1}{x^{\dfrac{1}{2}}+1}\ge-1\)
"=" khi x=1
B2:
a)\(A=x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2+4\left(x-y\right)-5\)
\(=\left(x-y\right)^2-1+4\left(x-y\right)-4\)
\(=\left(x-y+1\right)\left(x-y-1\right)+4\left(x-y-1\right)\)
\(=\left(x-y+5\right)\left(x-y-1\right)\)
b)\(P=x^4+2x^3+3x^2+2x+1\)
\(=\left(x^4+2x^3+x^2\right)+2\left(x^2+x\right)+1\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)
\(=\left(x^2+x+1\right)^2\ge0\forall x\)
Vậy MinP=0
c)\(Q=x^6+2x^5+2x^4+2x^3+2x^2+2x+1\)
\(=\left(x^2+x-1\right)\left(x^4+x^3+2x^2+x+3\right)+4\)
\(=\left(1-1\right)\left(x^4+x^3+2x^2+x+3\right)+4\)
\(=0\left(x^4+x^3+2x^2+x+3\right)+4=4\)
Vậy x^2+x=1 thì Q=4
B3:a)\(2xy+x+y=83\)
\(\Leftrightarrow x\left(2y+1\right)+\dfrac{1}{2}\left(2y+1\right)=\dfrac{167}{2}\)
\(\Leftrightarrow2x\left(2y+1\right)+1\left(2y+1\right)=167\)
\(\Leftrightarrow\left(2x+1\right)\left(2y+1\right)=167\)
Mà \(Ư\left(167\right)=\left\{\pm1;\pm167\right\}\)
\(\Leftrightarrow\left(x;y\right)=\left(-84;-1\right);\left(-1;-84\right);\left(0;83\right);\left(83;0\right)\)
Vậy...
b)\(y^2+2xy-3x-2=0\)
\(\Leftrightarrow x^2+y^2+2xy-x^2-3x-2=0\)
\(\Leftrightarrow\left(x+y\right)^2=x^2+3x+2\)
\(\Leftrightarrow\left(x+y\right)^2=\left(x+1\right)\left(x+2\right)\)
Vì \(x;y\in Z\) nên VT là số chính phương VP là tích 2 số nguyên liên tiếp
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2\end{matrix}\right.\)
Vậy...
B5:\(B=\dfrac{x^2+x+1}{x^2-x+1}\)
\(\Leftrightarrow x^2\left(B-1\right)+x\left(-B-1\right)+\left(B-1\right)=0\)
\(\Delta=\left(-B-1\right)^2-4\left(B-1\right)\left(B-1\right)\)
\(=-\left(B-3\right)\left(3B-1\right)\)
pt có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow\left(B-3\right)\left(3B-1\right)\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}B-3\le0\\3B-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}B\le3\\B\ge\dfrac{1}{3}\end{matrix}\right.\)
Min B=1/3 khi x=-1; Max B=3 khi x=1
Bài 1:
a)
\(\sin ^2x+\sin ^2x\cot^2x=\sin ^2x(1+\cot^2x)=\sin ^2x(1+\frac{\cos ^2x}{\sin ^2x})\)
\(=\sin ^2x.\frac{\sin ^2x+\cos^2x}{\sin ^2x}=\sin ^2x+\cos^2x=1\)
b)
\((1-\tan ^2x)\cot^2x+1-\cot^2x\)
\(=\cot^2x(1-\tan^2x-1)+1=\cot^2x(-\tan ^2x)+1=-(\tan x\cot x)^2+1\)
\(=-1^2+1=0\)
c)
\(\sin ^2x\tan x+\cos^2x\cot x+2\sin x\cos x=\sin ^2x.\frac{\sin x}{\cos x}+\cos ^2x.\frac{\cos x}{\sin x}+2\sin x\cos x\)
\(=\frac{\sin ^3x}{\cos x}+\frac{\cos ^3x}{\sin x}+2\sin x\cos x=\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin x\cos x}=\frac{(\sin ^2x+\cos ^2x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}\)
\(=\frac{1}{\frac{\sin 2x}{2}}=\frac{2}{\sin 2x}\)
Bài 2:
Áp dụng BĐT Cauchy Schwarz ta có:
\(P=\frac{a^2}{\sqrt{a(2c+a+b)}}+\frac{b^2}{\sqrt{b(2a+b+c)}}+\frac{c^2}{\sqrt{c(2b+c+a)}}\)
\(\geq \frac{(a+b+c)^2}{\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}}(*)\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\((\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq (a+b+c)(2c+a+b+2a+b+c+2b+c+a)\)
\(\Leftrightarrow (\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq 4(a+b+c)^2\)
\(\Rightarrow \sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}\leq 2(a+b+c)(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)
Dấu "=" xảy ra khi $a=b=c=1$
Bài 1:
Ta có:
\(\text{VT}=\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\)
\(=a-\frac{2ab^2}{a+2b^2}+b-\frac{2bc^2}{b+2c^2}+c-\frac{2ca^2}{c+2a^2}=(a+b+c)-2\left(\frac{ab^2}{a+2b^2}+\frac{bc^2}{b+2c^2}+\frac{ca^2}{c+2a^2}\right)\)
\(=3-2M(*)\)
Áp dụng BĐT Cauchy ta có:
\(M=\frac{ab^2}{a+b^2+b^2}+\frac{bc^2}{b+c^2+c^2}+\frac{ca^2}{c+a^2+a^2}\leq \frac{ab^2}{3\sqrt[3]{ab^4}}+\frac{bc^2}{3\sqrt[3]{bc^4}}+\frac{ca^2}{3\sqrt[3]{ca^4}}\)
\(\Leftrightarrow M\leq \frac{1}{3}(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2})\)
Tiếp tục áp dụng BĐT Cauchy:
\(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2}\leq \frac{ab+ab+1}{3}+\frac{bc+bc+1}{3}+\frac{ca+ca+1}{3}=\frac{2(ab+bc+ac)+3}{3}\)
Mà \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=3\) (quen thuộc)
\(\Rightarrow M\leq \frac{1}{3}.\frac{2.3+3}{3}=1(**)\)
Từ \((*);(**)\Rightarrow \text{VT}\geq 3-2.1=1\)
(đpcm)
Dấu bằng xảy ra khi $a=b=c=1$
Bài 2:
Áp dụng BĐT Cauchy -Schwarz:
\(\text{VT}=\frac{a^3}{a^2+a^2b^2}+\frac{b^3}{b^2+b^2c^2}+\frac{c^3}{c^2+a^2c^2}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{a^2+a^2b^2+b^2+b^2c^2+c^2+c^2a^2}\)
hay:
\(\text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+a^2b^2+b^2c^2+c^2a^2}(*)\)
Mặt khác, theo BĐT Cauchy ta dễ thấy:
\(a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2\)
\(\Rightarrow (a^2+b^2+c^2)^2\geq 3(a^2b^2+b^2c^2+c^2a^2)\)
\(\Leftrightarrow 1\geq 3(a^2b^2+b^2c^2+c^2a^2)\Rightarrow a^2b^2+b^2c^2+c^2a^2\leq \frac{1}{3}(**)\)
Từ \((*);(**)\Rightarrow \text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+\frac{1}{3}}=\frac{3}{4}(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)