5.1.a
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\Leftrightarrow\frac{4}{x+y}\le\frac{x+y}{xy}\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(x=y\)
b. \(\frac{ab}{c+1}=\frac{ab}{c+a+b+c}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
Tương tự: \(\frac{bc}{a+1}\le\frac{1}{4}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right)\) ; \(\frac{ca}{b+1}\le\frac{1}{4}\left(\frac{ca}{a+b}+\frac{ca}{b+c}\right)\)
Cộng vế với vế:
\(VT\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{bc}{a+c}+\frac{ab}{b+c}+\frac{ca}{b+c}+\frac{bc}{a+b}+\frac{ca}{a+b}\right)=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
5.2.a
Giống hệt 5.1.a nên ko chứng minh lại
b. \(M=\frac{1}{ab}+\frac{1}{a^2+b^2}=\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{1}{2ab}\)
\(M\ge\frac{4}{2ab+a^2+b^2}+\frac{1}{\frac{2\left(a+b\right)^2}{4}}=\frac{4}{\left(a+b\right)^2}+\frac{2}{\left(a+b\right)^2}=\frac{6}{\left(a+b\right)^2}=6\)
\(\Rightarrow M_{min}=6\) khi \(a=b=\frac{1}{2}\)
5.3.a
\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b=c\)
b/ Áp dụng câu trên:
\(\left(\frac{2xy}{x^2+y^2}\right)^2+\left(\frac{x}{y}\right)^2+\left(\frac{y}{x}\right)^2\ge\frac{2x^2}{x^2+y^2}+\frac{x.y}{y.x}+\frac{2y^2}{x^2+y^2}=\frac{2\left(x^2+y^2\right)}{x^2+y^2}+1=3\)
Dấu "=" xảy ra khi \(x=y\)
5.4.a
\(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)
\(\Leftrightarrow3a^2-3ab+3b^2\ge a^2+ab+b^2\)
\(\Leftrightarrow2\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow2\left(a-b\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b\)
b/ \(\frac{a^3}{a^2+ab+b^2}=a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\frac{ab\left(a+b\right)}{3\sqrt[3]{a^2.ab.b^2}}=a-\frac{1}{3}\left(a+b\right)=\frac{2a}{3}-\frac{b}{3}\)
Tương tự: \(\frac{b^3}{b^2+bc+c^2}\ge\frac{2b}{3}-\frac{c}{3}\) ; \(\frac{c^3}{c^2+ca+a^2}\ge\frac{2c}{3}-\frac{a}{3}\)
Cộng vế với vế:
\(VT\ge\frac{2}{3}\left(a+b+c\right)-\frac{1}{3}\left(a+b+c\right)=\frac{a+b+c}{3}\)
Dấu "=" xảy ra khi \(a=b=c\)
5.5.a
Giống hệt 5.1.a, chỉ thay x;y bằng a;b
b/ \(P=\frac{1}{4x^2+2}+\frac{1}{4y^2+2}+\frac{2}{xy}\)
\(P=\frac{1}{4x^2+2}+\frac{1}{4y^2+2}+\frac{1}{6xy}+\frac{1}{6xy}+\frac{5}{3xy}\)
\(P\ge\frac{16}{4x^2+2+4y^2+2+6xy+6xy}+\frac{5}{\frac{3\left(x+y\right)^2}{4}}\)
\(P\ge\frac{16}{4\left(x+y\right)^2+4+4xy}+\frac{20}{3\left(x+y\right)^2}\)
\(P\ge\frac{16}{4\left(x+y\right)^2+4+\left(x+y\right)^2}+\frac{20}{3\left(x+y\right)^2}=\frac{16}{4.2^2+4+2^2}+\frac{20}{3.2^2}=\frac{7}{3}\)
Dấu "=" xảy ra khi \(x=y=1\)
5.6.a
\(a+b\ge2\sqrt{ab}\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b\)
b/ \(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)
Tương tự: \(\frac{b^2}{c+a}+\frac{c+a}{4}\ge b\) ; \(\frac{c^2}{a+b}+\frac{a+b}{4}\ge c\)
Cộng vế với vế:
\(P+\frac{a+b+c}{2}\ge a+b+c\Rightarrow P\ge\frac{a+b+c}{2}=\frac{6}{2}=3\)
Dấu "=" xảy ra khi \(a=b=c=2\)
5.7.a
\(\frac{x}{\sqrt{x-1}}\ge2\) \(\Leftrightarrow x\ge2\sqrt{x-1}\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+1\ge0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(x=2\)
b/ Đặt \(P=\frac{a^2}{b-1}+\frac{b^2}{a-1}\ge2\sqrt{\frac{a^2b^2}{\left(a-1\right)\left(b-1\right)}}=2.\frac{a}{\sqrt{a-1}}.\frac{b}{\sqrt{b-1}}\)
Mà theo chứng minh câu a ta có:
\(\frac{a}{\sqrt{a-1}}\ge2\) ; \(\frac{b}{\sqrt{b-1}}\ge2\Rightarrow P\ge2.2.2=8\)
Dấu "=" xảy ra khi \(a=b=2\)
lm hộ mk vs
, biết rằng n chia hết cho 99.