Eigenvalue and spectral condition 
Let $A=\begin{pmatrix} 1& 1 \\ a^2 &1 \end{pmatrix} \text{ with } a\in (0,\frac{1}{2}]$. Show $$cond_2(A)=||A||_2 \cdot ||A^{-1}||_2\leq 4(1-a^2)^{-1}$$ by first showing $||A||^2_2\leq||A||_1||A||_{\infty}$.

$||A||^2_2$ is the maximal eigenvalue of $A^TA$ and $||A||_1||A||_{\infty}=2\cdot2 = 4$. Prove $||A||^2_2\leq||A||_1||A||_{\infty}=4$ by contradiction: Suppose $\lambda_{max} >4$, then $$4<\lambda_{\ast}=\frac{\sqrt{a^8+2a^4+4a^2+1}+a^4+3}{2} \iff 5<\sqrt{a^8+\underbrace{2a^4}_{\leq \frac{1}{2}}+\underbrace{4a^2}_{\leq 1}+1}+\underbrace{a^4}_{\leq 1} \\ \leq \sqrt{4}+1 = 3$$ contradiction!
So I know $||A||^2_2\leq 4 \Rightarrow ||A||_2\leq 2$ but I need to show $cond_2(A) = \underbrace{||A||_2}_{\leq 2}||A^{-1}||_2\leq 4(1-a^2)^{-1}$. I don't know why I needed to show $||A||^2_2\leq||A||_1||A||_{\infty}$ first? How does it help? I still need the maximal eigenvalue of $(A^{-1})^TA^{-1}$. The eigenvalues are $$\lambda_1=\frac{\sqrt{a^4-2a^2+5}-a^2+1}{2\sqrt{a^4-2a^2+5}-4} \\ \lambda_2=\frac{\sqrt{a^4-2a^2+5}-a^2-1}{2\sqrt{a^4-2a^2+5}+4}$$ What am I supposed to do now? Thanks for any help!

$0 \lt  1-a^2 = \det\big(A)=\lambda_1\cdot \lambda_2 = \sigma_1\cdot \sigma_2$
i.e. the product of singular values gives the modulus of the determinant (and in particular when the determinant positive it is equal to the product of singular values)
proof: using polar decomposition
$\det\big(A\big) =\det\big(UP\big)=\det\big(U\big)\det\big(P\big)=\pm \det\big(P\big) = \pm(\sigma_1\cdot\sigma_2)$
now I assume you know that $||A||_2 = \sigma_1$ and $||A^{-1}||_2=\frac{1}{\sigma_2}$ so we may write
$\text{cond}_2(A)=||A||_2 \cdot ||A^{-1}||_2 = \frac{\sigma_1}{\sigma_2}=\frac{\sigma_1^2}{\sigma_1\sigma_2}\lt \frac{\sigma_1^2+\sigma_2^2}{\sigma_1\sigma_2}=\frac{\big \Vert A\big \Vert_F^2}{\det\big(A\big)}= \frac{3+a^4}{1-a^2}\lt \frac{4}{1-a^2}$
