Áp dụng BĐT cosi:

\(a\sqrt{1-b^2}=\sqrt{a^2\left(1-b^2\right)}\le\dfrac{a^2+1-b^2}{2}\)

Tương tự cx có: \(b\sqrt{1-c^2}\le\dfrac{b^2+1-c^2}{2}\)

\(c\sqrt{1-a^2}\le\dfrac{c^2+1-a^2}{2}\)

Cộng vế với vế \(\Rightarrow VT\le\dfrac{3}{2}\)

Dấu = xảy ra <=> \(\left\{{}\begin{matrix}a^2=1-b^2\\b^2=1-c^2\\c^2=1-a^2\end{matrix}\right.\) \(\Leftrightarrow a^2+b^2+c^2=3-\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow a^2+b^2+c^2=\dfrac{3}{2}\) (đpcm)

\(\dfrac{a}{a^4+b^2}+\dfrac{b}{a^2+b^4}\le\dfrac{a}{2\sqrt{a^4b^2}}+\dfrac{b}{2\sqrt{a^2b^4}}=\dfrac{a}{2a^2b}+\dfrac{b}{2ab^2}=\dfrac{1}{ab}=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=1\)

Ta có: \(0\le a\le b\le c\le1\Leftrightarrow\left\{{}\begin{matrix}1-a\ge0\\1-b\ge0\end{matrix}\right.\Leftrightarrow\left(1-a\right)\left(1-b\right)\ge0\)

\(\Rightarrow1-b-a+ab\ge0\Leftrightarrow1+ab\ge a+b\)(1)

Tiếp tục chứng minh ta được: \(0\le a\le b\le c\le1\Leftrightarrow\left\{{}\begin{matrix}1\ge c\\ab\ge0\end{matrix}\right.\)(2)

Cộng theo vế pt(1) với pt(2) ta được:

\(1+ab+1+ab\ge a+b+c+0\)

\(\Rightarrow2\left(ab+1\right)\ge a+b+c\)

Nên: \(\dfrac{c}{ab+1}=\dfrac{2c}{2\left(ab+1\right)}\le\dfrac{2c}{a+b+c}\)

Chứng minh tương tự suy ra đpcm

Câu hỏi của Phạm Quốc Anh - Toán lớp 7 - Học toán với OnlineMath

Ta có đánh giá sau với a không âm:

\(\dfrac{a}{1+a^2}\le\dfrac{36a+3}{50}\)

Thật vậy, BĐT tương đương:

\(\left(36a+3\right)\left(a^2+1\right)\ge50a\)

\(\Leftrightarrow\left(3a-1\right)^2\left(4a+3\right)\ge0\) (luôn đúng)

Tương tự: \(\dfrac{b}{1+b^2}\le\dfrac{36b+3}{50}\) ; \(\dfrac{c}{1+c^2}\le\dfrac{36c+3}{50}\)

Cộng vế: \(VT\le\dfrac{36\left(a+b+c\right)+9}{50}=\dfrac{9}{10}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Ta chứng minh bđt phụ \(\dfrac{a}{1+a^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\)

Thật vậy bđt trên \(\Leftrightarrow\dfrac{-3a^2+10a-3}{10\left(1+a^2\right)}-\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\le0\)

\(\Leftrightarrow\left(a-\dfrac{1}{3}\right)\left[\dfrac{3\left(3-a\right)}{10\left(1+a^2\right)}-\dfrac{18}{25}\right]\le0\)

\(\Leftrightarrow-\dfrac{36\left(a-\dfrac{1}{3}\right)^2\left(\dfrac{3}{4}+a\right)}{50\left(1+a^2\right)}\le0\) ( luôn đúng với mọi \(a\)\(\ge\)0)

Tương tự cũng có:\(\dfrac{b}{1+b^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(b-\dfrac{1}{3}\right)\); \(\dfrac{c}{1+c^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(c-\dfrac{1}{3}\right)\)

Cộng vế với vế => VT\(\le\dfrac{9}{10}+\dfrac{18}{25}\left(a+b+c-1\right)=\dfrac{9}{10}\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{3}\)

\(ab+1\le b\Rightarrow a+\dfrac{1}{b}\le1\)

Đặt \(\left(a;\dfrac{1}{b}\right)=\left(x;y\right)\Rightarrow x+y\le1\)

\(P=x+\dfrac{1}{x^2}+y+\dfrac{1}{y^2}=\left(\dfrac{x}{2}+\dfrac{x}{2}+\dfrac{1}{16x^2}\right)+\left(\dfrac{y}{2}+\dfrac{y}{2}+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)

\(P\ge3\sqrt[3]{\dfrac{x^2}{64x^2}}+3\sqrt[3]{\dfrac{y^2}{64y^2}}+\dfrac{15}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(P\ge\dfrac{3}{2}+\dfrac{15}{32}\left(\dfrac{4}{x+y}\right)^2\ge\dfrac{3}{2}+\dfrac{15}{32}.\left(\dfrac{4}{1}\right)^2=9\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\) hay \(\left(a;b\right)=\left(\dfrac{1}{2};2\right)\)

Áp dụng bất đẳng thức AM - GM ta có:

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

\(=\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ca}}\le\dfrac{1}{\sqrt{a^2}}+\dfrac{1}{\sqrt{b^2}}+\dfrac{1}{\sqrt{c^2}}\)

\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

Dấu " = " xảy ra khi \(a=b=c\)

Vậy...

Áp dụng BĐT \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\). Tương tự cho 2 BĐT còn lại có:

\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c};\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+c}\)

Cộng theo vế 3 BĐT trên ta có:

\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)

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

Đẳng thức xảy ra khi \(a=b=c\)

\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) Chứng minh tương tự ta được:

\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+a}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\)

\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+a}+\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\left(1+1+1\right)=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)

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

Tương tự: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)

Cộng vế:

\(VT\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Đặt \(T=\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\)

\(BDT\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ca}{c+a}-b+\dfrac{c^2+ab}{a+b}-c\ge0\)

\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-ab-bc}{a+c}+\dfrac{c^2+ab-ac-bc}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{b+c}+\dfrac{\left(b-a\right)\left(b-c\right)}{a+c}+\dfrac{\left(c-a\right)\left(c-b\right)}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)\left(a^2-c^2\right)+\left(b^2-a^2\right)\left(b^2-c^2\right)+\left(c^2-a^2\right)\left(c^2-b^2\right)}{T}\ge0\)

\(\Leftrightarrow\dfrac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{T}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2}{2T}\ge0\)

Xảy ra khi \(a=b=c\)

\(BĐT\Leftrightarrow\sum\left(\dfrac{1}{a}-\dfrac{b+c}{a^2+bc}\right)\ge0\)

\(\Leftrightarrow\sum\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\)

Giả sử \(a\ge b\ge c\)thì

\(\dfrac{\left(a-b\right)\left(a-c\right)}{a\left(a^2+bc\right)}\ge0\).vậy nên chỉ cần chứng minh

\(\dfrac{\left(b-c\right)\left(b-a\right)}{b\left(b^2+ac\right)}+\dfrac{\left(c-a\right)\left(c-b\right)}{c\left(c^2+ab\right)}\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\dfrac{b-a}{b\left(b^2+ac\right)}+\dfrac{a-c}{c\left(c^2+ab\right)}\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)\left[\left(b-a\right)\left(c^3+abc\right)+\left(a-c\right)\left(b^3+abc\right)\right]\ge0\)

\(\Leftrightarrow\left(b-c\right)^2\left(b+c\right)\left(ab+ac-bc\right)\ge0\)( đúng vì \(a\ge b\ge c\))

Vậy BĐT được chứng minh.

Dấu = xảy ra khi a=b=c