\(S=\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\)

\(S=\left(1+\frac{1}{1-b}\right)\left(1+\frac{1}{1-a}\right)\)

\(S=\frac{1-b+1}{1-b}\times\frac{1-a+1}{1-a}\)

\(S=\frac{\left(2-b\right)\left(2-a\right)}{\left(1-b\right)\left(1-a\right)}\)

\(S=\frac{4-2a-2b+ab}{1-a-b+ab}=\frac{4-2\left(a+b\right)+ab}{1-\left(a+b\right)+ab}\)

\(S=\frac{4-2+ab}{1-1+ab}=\frac{2+ab}{ab}=1+\frac{2}{ab}\)(*)

 từ \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2-2ab\ge0\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow4ab\le1\Leftrightarrow ab\le\frac{1}{4}\Leftrightarrow\frac{1}{ab}\ge4\)

\(\Leftrightarrow\frac{2}{ab}\ge8\)(1)

thay (1) vào (*) có

\(S=1+\frac{2}{ab}\ge1+8=9\)

vậy GTNN của \(S=9\Leftrightarrow x=y=\frac{1}{2}\)

Cảm ơn bạn vì đã giúp đỡ mình! Thanks very much!

P=2/(a^2+b^2)+2/2ab+68/2ab. ap dung bdt 1/a+1/b>=4/a+b. ta co 2/(a^2+b^2)+2/2ab>=

Có: \(\frac{a}{b+c+d}+\frac{b+c+d}{a}=\frac{a}{b+c+d}+\frac{b+c+d}{9a}+\frac{8\left(b+c+d\right)}{9a}\)

\(\ge2\sqrt{\frac{a}{b+c+d}.\frac{b+c+d}{9a}}+\frac{8\left(b+c+d\right)}{9a}\)

\(=\frac{2}{3}+\frac{8\left(b+c+d\right)}{9a}\)

Tương tự ba BĐT còn lại và cộng theo vế thu được:

\(\Sigma_{cyc}\left(\frac{a}{b+c+d}+\frac{b+c+d}{a}\right)=\frac{8}{3}+\frac{8}{9}\left(\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+c}{c}+\frac{a+b+c}{d}\right)\)

\(\ge\frac{8}{3}+\frac{32}{9}\sqrt[4]{\frac{\left(b+c+d\right)\left(c+d+a\right)\left(d+a+c\right)\left(a+b+c\right)}{abcd}}\)

\(\ge\frac{8}{3}+\frac{32}{9}\sqrt[4]{\frac{3^4.abcd}{abcd}}=\frac{40}{3}\)

Đẳng thức xảy ra khi a = b =c = d

P/s: Tính sai chỗ nào tự sửa nhá, dạo này hay nhầm lắm!

Vì \(a;b>0\) nên \(\frac{a+b}{\sqrt{a+b}}>0;\frac{\sqrt{a+b}}{a+b}>0\)

Áp dụng bđt AM - GM ta có :

\(S=\frac{a+b}{\sqrt{a+b}}+\frac{\sqrt{a+b}}{a+b}\ge2\sqrt{\frac{a+b}{\sqrt{a+b}}.\frac{\sqrt{a+b}}{a+b}}=2.\sqrt{1}=2\)

Dấu "=" xảy ra <=> \(\frac{a+b}{\sqrt{a+b}}=\frac{\sqrt{a+b}}{a+b}\Leftrightarrow a+b=1\)

Vậy GTNN của S là 2 tại a + b = 1

By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

Ta co:

\(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2=\left(\frac{1}{a}-2\right)^2+\left(\frac{1}{b}-2\right)^2+6\left(\frac{1}{a}+\frac{1}{b}\right)-6\ge\frac{24}{a+b}-6=18\)

Dau '=' xay ra khi \(a=b=\frac{1}{2}\)