§1. Bất đẳng thức

Lời giải:

Ta sẽ CM BĐT trung gian sau:

\(P\geq \frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)

\(\Leftrightarrow x^2\left ( \frac{1}{y+z}-\frac{1}{x+y} \right )+y^2\left ( \frac{1}{x+z}-\frac{1}{z+y} \right )+z^2\left ( \frac{1}{x+y}-\frac{1}{z+x} \right )\geq 0\)

\(\Leftrightarrow x^2(x^2-z^2)+y^2(y^2-x^2)+z^2(z^2-y^2)\geq 0\)

\(\Leftrightarrow (x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2\geq 0\) (luôn đúng)

Giờ ta sẽ tìm min \(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)

Hiển nhiên \(\sum \frac{x^2}{x+y}=\sum \frac{y^2}{x+y}\) nên

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

Áp dụng BĐT Cauchy-Schwarz:

\(A\geq \frac{1}{2}\frac{(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2}{2(x+y+z)}=\frac{9}{x+y+z}\)

Áp dụng BĐT Cauchy: \(\sqrt{x^2+y^2}\geq \frac{x+y}{\sqrt{2}}\)

Tương tự với các số còn lại suy ra \(6\geq \sqrt{2}.(x+y+z)\Rightarrow x+y+z\leq 3\sqrt{2}\)

\(\Rightarrow A\geq \frac{3\sqrt{2}}{2}\) kéo theo \(P_{\min}=\frac{3\sqrt{2}}{2}\)

Dấu bằng xảy ra khi \(x=y=z=\sqrt{2}\)

Em làm đại ạ ; có sai sót mong anh chị bỏ qua ạ !!

\(S=x+y+\dfrac{1}{x}+\dfrac{1}{y}\\ =\left(x+\dfrac{4}{9x}\right)+\left(y+\dfrac{4}{9y}\right)+\dfrac{5}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\\ \ge2.\sqrt{x.\dfrac{4}{9x}}+2.\sqrt{y.\dfrac{4}{9y}}+\dfrac{5}{9}.\dfrac{\left(1+1\right)^2}{x+y}\\ =\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{5}{9}.\dfrac{4}{x+y}\\ =\dfrac{8}{3}+\dfrac{20}{9\left(x+y\right)}\\ x+y\le\dfrac{4}{3}\\ \Leftrightarrow9\left(x+y\right)\le12\\ \Leftrightarrow\dfrac{20}{9\left(x+y\right)}\ge\dfrac{20}{12}=\dfrac{5}{3}\\ \Leftrightarrow S\ge\dfrac{8}{3}+\dfrac{5}{3}=\dfrac{13}{3}\)

/Dấu = xảy ra khi x=y=2/3

\(P=\left(a^2+4a+12\right)+\left(\dfrac{36a+81}{a^2}+3\right)\)

\(=\left(a+1\right)\left(a+3\right)+\dfrac{3\left(a+9\right)\left(a+3\right)}{a^2}+9\)

\(=\left(a+3\right)\left(\left(a+1\right)+\dfrac{3\left(a+9\right)}{a^2}\right)+9\)

\(=\left(a+3\right)^2\left(a^2-2a+9\right)+9\ge9\)

\("="\Leftrightarrow a=-3\)

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*

Ta có \(\dfrac{a^2}{a+b^2}=a-\dfrac{ab^2}{a+b^2}\ge a-\dfrac{ab^2}{2b\sqrt{a}}=a-\dfrac{ab}{2\sqrt{a}}\)

Thiết lập tương tự và thu lại ta có :

\(VT\ge3-\left(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\right)\)

Xét \(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}=\sqrt{\dfrac{a^2b^2}{4a}}+\sqrt{\dfrac{b^2c^2}{4b}}+\sqrt{\dfrac{a^2c^2}{4c}}\)

Áp dụng bđt Cauchy ta có \(\sqrt{\dfrac{a^2b^2}{4a}}=\sqrt{\dfrac{ab}{2a}.\dfrac{ab}{2}}\le\dfrac{\dfrac{b}{2}+\dfrac{ab}{2}}{2}\)

Thiết lập tương tự và thu lại ta có :

\(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\le\dfrac{\dfrac{a+b+c}{2}+\dfrac{ab+bc+ac}{2}}{2}=\dfrac{\dfrac{3}{2}+\dfrac{ab+bc+ac}{2}}{2}\left(1\right)\)

Theo hệ quả của bđt Cauchy ta có \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

\(\Rightarrow ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\)

\(\Rightarrow\dfrac{\dfrac{3}{2}+\dfrac{ab+bc+ac}{2}}{2}\le\dfrac{\dfrac{3}{2}+\dfrac{3}{2}}{2}=\dfrac{3}{2}\left(2\right)\)

Từ ( 1 ) và ( 2 ) ta có \(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\le\dfrac{3}{2}\)

\(\Rightarrow3-\left(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)

\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)

Dấu '' = '' xảy ra khi \(a=b=c=1\)

Ta luôn có \(\left(\dfrac{1}{\sqrt{a}}-\dfrac{1}{\sqrt{b}}\right)^2\ge0\forall a;b\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\)

\(\Leftrightarrow2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{2}{\sqrt{ab}}+\dfrac{1}{a}+\dfrac{1}{b}\)

\(\Leftrightarrow\dfrac{2\left(a+b\right)}{ab}\ge\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\right)^2\)

\(\Leftrightarrow\sqrt{\dfrac{2\left(a+b\right)}{ab}}\ge\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(\sqrt{2}\left(\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{a+c}{ac}}\right)\ge2\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)\)

\(\Leftrightarrow\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{a+c}{ac}}\ge\sqrt{\dfrac{2}{a}}+\sqrt{\dfrac{2}{b}}+\sqrt{\dfrac{2}{c}}\)

\("="\Leftrightarrow a=b=c\)

a) Sai với \(a=1,b=2\)

b)

Thực hiện biến đổi tương đương:

\(\frac{a}{3b}+\frac{b(a+b)}{a^2+ab+b^2}\geq 1\)

\(\Leftrightarrow \frac{a}{3b}+\frac{b(a+b)+a^2}{a^2+ab+b^2}-\frac{a^2}{a^2+ab+b^2}\geq 1\)

\(\Leftrightarrow \frac{a}{3b}-\frac{a^2}{a^2+ab+b^2}\geq 0\)

\(\Leftrightarrow \frac{1}{3b}-\frac{a}{a^2+ab+b^2}\geq 0\)

\(\Leftrightarrow \frac{a^2+ab+b^2-3ab}{3b(a^2+ab+b^2)}\geq 0\)

\(\Leftrightarrow \frac{(a-b)^2}{3b(a^2+ab+b^2)}\geq 0\) (luôn đúng)

Do đó ta có đpcm. Dấu bằng xảy ra khi $a=b$

c) BĐT sai với \(a=1,b=2\)