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\(3a^2+4ab+b^2=3a^2+3ab+ab+b^2=3a\left(a+b\right)+b\left(a+b\right)=\left(3a+b\right)\left(a+b\right)\)
xong AM -GM
Ta có: \(1=x+y\ge2\sqrt{xy}\)
\(\Rightarrow4xy\le1\)
\(S=\frac{1}{x^2+y^2}+\frac{3}{4xy}\)
\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{4xy}\)
\(\ge\frac{4}{x^2+y^2+2xy}+\frac{1}{1}=\frac{4}{\left(x+y\right)^2}+1=\frac{4}{1}+1=5\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
Áp dụng BĐT AM - MG ta có :
\(xy\)\(\le\)\(\frac{\left(x+y\right)^2}{4}\)\(=\)\(\frac{1}{4}\)
Áp dụng BĐT Cauchy - Schwarz dạng Engel :
\(S\)\(=\)\(\frac{1}{x^2+y^2}\)\(-\)\(\frac{3}{4xy}\)\(=\)\(\frac{1}{x^2+y^2}\)\(-\)\(\frac{2}{4xy}\)\(-\)\(\frac{1}{4xy}\)
\(=\)\(\frac{1}{x^2+y^2}\)\(-\)\(\frac{1}{2xy}\)\(-\)\(\frac{1}{4xy}\)\(\ge\)\(\frac{\left(1-1\right)^2}{x^2-y^2-2xy}\)\(-\)\(\frac{1}{4xy}\)
\(\ge\)\(\frac{\left(1+1\right)^2}{\left(x+y\right)^2}\)\(-\)\(\frac{1}{4.\frac{1}{4}}\)\(=\)\(4\)\(-\)\(1\)\(=\)\(5\)
Xảy ra khi \(x\)\(=\)\(y\)\(=\)\(\frac{1}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{\sqrt{3a^2+4ab+b^2}}=\dfrac{1}{\sqrt{\left(a+b\right)\left(3a+b\right)}}=\dfrac{\sqrt{2}}{\sqrt{\left(2a+2b\right)\left(3a+b\right)}}\)
\(\ge\dfrac{\sqrt{2}}{\dfrac{2a+2b+3a+b}{2}}=\dfrac{\sqrt{2}}{\dfrac{5a+3b}{2}}=\dfrac{2\sqrt{2}}{5a+3b}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{1}{\sqrt{3b^2+4bc+c^2}}\ge\dfrac{2\sqrt{2}}{5b+3c};\dfrac{1}{\sqrt{3c^2+4ca+a^2}}\ge\dfrac{2\sqrt{2}}{5c+3a}\)
Cộng theo vế 3 BĐT trên ta có:
\(P\ge\dfrac{2\sqrt{2}}{5a+3b}+\dfrac{2\sqrt{2}}{5b+3c}+\dfrac{2\sqrt{2}}{5c+3a}\)
\(\ge\dfrac{18\sqrt{2}}{8\left(a+b+c\right)}=\dfrac{18\sqrt{2}}{8}=\dfrac{9\sqrt{2}}{4}\)
Xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Dùng bđt AM - GM cho 7 số; 2 số và 3 số không âm, ta được:
\(a^3c^2+a^3c^2+a^3c^2+b^3a^2+b^3a^2+1+1\ge7a\)(1)
\(b^3a^2+b^3a^2+b^3a^2+c^3b^2+c^3b^2+1+1\ge7b\)(2)
\(c^3b^2+c^3b^2+c^3b^2+a^3c^2+a^3c^2+1+1\ge7c\)(3)
\(\frac{a+b+c}{2}+\frac{9}{2\left(a+b+c\right)}\ge3\)
\(a+b+c\ge3\)
Từ (1); (2); (3) suy ra \(a^3c^2+b^3a^2+c^3b^2\ge\frac{7\left(a+b+c\right)}{5}-\frac{6}{5}\)
\(P=\text{Σ}_{cyc}\frac{a}{b^2}+\frac{9}{2\left(a+b+c\right)}=\text{Σ}_{cyc}a^3c^2+\frac{9}{2\left(a+b+c\right)}\)
\(\ge\frac{7\left(a+b+c\right)}{5}+\frac{9}{2\left(a+b+c\right)}-\frac{6}{5}\)
\(=\frac{a+b+c}{2}+\frac{9}{2\left(a+b+c\right)}+\frac{9\left(a+b+c\right)}{10}-\frac{6}{5}\)
\(\ge3+\frac{9}{10}.3-\frac{6}{5}=\frac{9}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
Áp dụng BĐT Cauchy ta có: \(xy+\frac{1}{xy}\ge2\sqrt{xy\cdot\frac{1}{xy}}=2\)
Vậy \(M\text{inS}=2\) với mọi \(x;y\ge1\)
\(\frac{a^2}{1+b}+\frac{1+b}{4}\ge a\)
\(\frac{b^2}{1+c}+\frac{1+c}{4}\ge b\)
\(\frac{c^2}{1+a}+\frac{1+a}{4}\ge c\)
=>\(A\ge a+b+c-\frac{1}{4}\left(3+a+b+c\right)=\frac{3}{4}\left(a+b+c-1\right)\ge\frac{3}{4}\left(3\sqrt[3]{abc}-1\right)=\frac{3}{2}\)
A min = 3/2 khi x= y =z =1